Measure Theory and Lebesgue Integration an introductory course

Written by: Isaac Solomon Prerequisites: A course in Real Analysis, covering Riemann/Riemann-Stieltjes integration.

Table of Contents 1. A review of the Riemann/Riemann-Stieltjes integration. The history of its development, its properties, and its shortcomings. 2. Lebesgue Measure Zero and a classification of the space of the Riemann-Integrable Functions. 3. Lebesgue Outer Measure. Measurable sets, Non-Measurable sets, and the Axiom of Choice. The Cantor-Lebesgue function. 4. Proving that the space of Measurable sets forms a σ-algebra containing the Borel sets. 5. Measurable functions, and the four-step construction of the Lebesgue integral. 6. Littlewood’s Three Principles. The Bounded and Uniform Convergence Theorems. Fatou’s Lemma and the Dominated and Monotone Convergence Theorems. 7.The Riesz-Fischer Theorem: L1 is complete.

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Measure Theory and Lebesgue Integration: Lesson I

“If only I had the theorems! Then I should find the proofs easily enough.” Bernard Riemann (1826-1866) A review of Riemann integration. The history of its development, its properties, and its shortcomings.

The History of the Riemann Integral There are few branches of mathematics whose origins are as contentious as that of calculus. Some say that it was the British mathematician Isaac Newton who first discovered calculus, others insist it was the German mathematician Gottfriend Liebniz, and still others are certain that Stephen Wolfram invented in 1988∗ . Nevertheless, the calculus they developed was very different from our own, for it was based on the existence of infinitesimals, units of length that were “infinitely small.” The problem with this infinitesimal calculus was that nobody could make the notion of “infinitesimal” rigorous. It was not until the turn of the 19th century that the mathematicians Augustin-Louis Cauchy and Karl Weierstrass eliminated the infinitesimal by constructing the formal  − δ notion of limit. With this rigorous groundwork already prepared for him, Bernhard Riemann finally developed the familiar Riemann Integral we use today. Note that although infinitesimals were abandoned in traditional calculus, there is a branch of math known as “Non-Standard Analysis” that uses set-theoretic ideas to make the notion of infinitesimals rigorous. *This is, of course, a joke. The reader will laugh now.

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The Construction of the Riemann Integral The Riemann integral, conceptually, is a way of finding the area under a curve. In other words, it is a way of measuring how much “information” or “mass” a function accrues over some interval. We shall construct it rigorously below. Let [a, b] be an interval. We define a partition P of [a, b] to be a finite collection of points P = {a = x0 < x1 < · · · < xn = b} contained in my interval. These points divides my interval [a, b] in a collection of sub-intervals [a = x0 , x1 ], [x1 , x2 ], [x2 , x3 ] · · · , [xn−2 , xn−1 ], [xn−1 , xn = b] Now, let us define the notion of a step function. A step function ψ is a 2-tuple consisting of a finite partition P with n + 1 elements, and a finite collection {a1 , · · · , an } of values, such that ψ is equal to ai on the interval (xi−1 , xi ). For example, let my interval [a, b] = [0, 10], and let my partition P = {0, 1, 2, · · · , 8, 9, 10}. Suppose that my set of values is {0, 1, 2, · · · , 7, 8, 9}. Then the associated step function ψ would look like

Now, let us define the integral of a step function ψ: Z

b

ψ= a

n X

ai (xi − xi−1 )

i=1

It is easy to see that this is the sum of the areas of the rectangles drawn beneath my step function. Now, let f be a bounded function on [a, b]. We define the Lower Riemann integral: (Z ) Z b

b

ψ | ψ is a step function, and ψ ≤ f on [a, b]

= sup a

a

and the Upper Riemann Integral: (Z Z b

ψ | ψ is a step function, and ψ ≥ f on [a, b]

= inf a

)

b

a

3

If b

Z

b

Z =

a

a

we say f ∈ R([a, b]), the space of Riemann integrable functions on [a, b], and Z

b

b

Z

a

Z

b

f=

= a

a

Properties of the Riemann Integral We know that if a bounded function is continuous, monotonic, or has only finitely many discontinuities, it is Riemann integrable. The proof of these statements can be found in any introductory analysis text, so I won’t cover it here. Next time, we shall give a precise “if-and-only-if” condition for determining when a function is Riemann integrable. Now, let us consider an interesting function that is not Riemann integrable, the Dirichlet Function. Define f on [0, 1] by setting f (x) = 1 when x is rational, and 0 when x is irrational. I don’t recommend you spend too long trying to imagine what the graph of this function looks like. Let ψ be a step function with partition P1 , such that ψ ≤ f . Since every interval contains both rational and irrational points, ψ is forced to equal zero on every interval. Thus (Z ) Z b

b

ψ | ψ is a step function, and ψ ≤ f on [a, b]

= sup a

=0

a

Conversely, let φ be a step function with partition P2 , such that φ ≥ f . Since every interval contains both rational and irrational points, φ must equal 1 on every interval, so that (Z ) Z b

b

φ | φ is a step function, and φ ≥ f on [a, b]

= inf a

=1

a

We see that the upper and lower Riemann integrals are different, and so f is not integrable. Another shortcoming of the Riemann integral is that it is not defined on infinite sets. To take the integral over R, one has to take a sequence of integrals on larger and larger intervals, which can often be hard to work with. These are called improper integrals. Lastly, let us define a norm on R([a, b]). If f, g ∈ R([a, b]), Z k f − g k=

b

|f − g| a

If k f − g k= 0, we say that f = g in the `1 norm. Now, let fn be a Cauchy sequence of Riemann integrable functions. It is not necessarily the case that they converge in R([a, b]). We therefore say that the space of Riemann integrable functions is not complete, and therefore not a Banach Space (complete, normed vector space). This sucks, because Banach spaces are some of the most wellstudied and understood spaces. Clearly, if we want to develop a theory of integration that is flexible and dynamic, we need to ditch the Riemann integral.

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Measure Theory and Lebesgue Integration: Lesson II

“In mathematics the art of proposing a question must be held of higher value than solving it.” Georg Cantor (1845-1918) Lebesgue Measure Zero and a classification of the space of the Riemann-Integrable Functions.

When does the Riemann integral exist? As we saw with the Dirichlet Function, there are some functions that just aren’t Riemann integrable. The question then becomes, “well, why not?” The Dirichlet function is rather peculiar, to be sure, and it’s not obvious precisely where integration fails. Are there functions that look like the Dirichlet function that are Riemann integrable? These and further questions will be explored in this lesson. Well, let’s start with what we know. If a function is continuous, it’s easy to imagine that I can approximate its area smoothly with rectangles from above and below. It’s also easy to prove. What if a function f has a single discontinuity? Is is still Riemann integrable? The answer is still yes. To see this, suppose that the point p is the discontinuity of our function. Let ψ be any step function approximating f from above or below, with partition P . We construct a sequence of new partitions 1 1 } ∪ {p + } n n As n gets larger, the interval around p gets smaller and smaller. Since f is bounded by some constant M , the total amount it can vary on any partition is 2M . As such, although the upper and lower rectangles on (p − n1 , p + n1 ) aren’t equal, that difference can be made as small as I like by shrinking the partition, and so f ∈ R. (Try making this rigorous!) Pn = P ∪ {p −

The above process can be mirrored when f has any finite number of discontinuities, and so such functions are also Riemann integrable. Well, what about a function with countably many discontinuities? Are these functions still in R? The answer, surprisingly, is still yes. At this point, those readers who know some set theory may be thinking to themselves, “Aha! Clearly, countably many discontinuities aren’t enough. But what if we had an uncountable number of discontinuities? There’s no way such a function can still be Riemann integrable!”. (Readers unfamiliar with countable/uncountable infinities should pause here and look them up on Wikipedia). Incredibly enough, there are functions with uncountably many discontinuities that are still Riemann integrable. In the exercises at the end of this lesson, I will outline examples to demonstrate these facts. However, it is rather challenging to prove those examples using only the machinery developed thus far. To that end, we will need to develop a perfunctory notion of measure. 5

Lebesgue Measure Zero As we have seen, we cannot tell if a function is Riemann integrable or not merely by counting its discontinuities. One possible alternative is to look at how much space the discontinuities take up. Our question then becomes: how can one tell, rigorously, how much space a set takes up? Is there a useful definition that will coincide with our intuitive understanding of volume or area? If we had been faced with this problem a little over a century ago, we might’ve shrugged our shoulders and gone on with our day. Luckily for us, in 1903, a brilliant French mathematician named Henri Lebesgue developed a way of assigning to each set a “size” or “measure.” It can constructed as follows. Let E be a subset of R. One might also write E ∈ 2R or E ∈ P (R), which says that E is an element of the power set of R, the collection of all subsets of R. Now, let {(ak , bk )}∞ k=1 be a collection of open intervals with finite endpoints. We say that this collection covers E if ∞ [

E⊂

(ak , bk )

k=1

To every open cover {(ak , bk )}∞ k=1 we can associate a length ∞ X

(bk − ak )

k=1

As one can imagine, there are infinitely many different open covers, some of which might be ”tighter” or ”looser” than others. To get the best fit, we define the Lebesgue outer measure ) (∞ ∞ [ X (ak , bk ) (bk − ak ) | E ⊂ |E| = inf k=1

k=1

This says: to find the outer measure of E, look at every possible open cover, and try to find the one with the smallest length. It’s as if E is a collection of smudges on my bathroom door, and I’m trying to cover it with as little paint as possible. That being said, E has Lebesgue measure zero if |E| = 0. What sets have Lebesgue measure zero? I think we can all agree that the empty set, ∅, should have Lebesgue measure zero. What about larger sets? Theorem: Any countable set has Lebesgue measure zero. Proof: Let  > 0, and let {qn } be an enumeration of my countable set E. Define the following open cover:   , qn + n )}∞ n=1 2n 2 The sum of the lengths of O is O = {(qn −

∞  X n=1

(qn +

∞ X     ) − (q − ) = 2 = 2 n n 2n 2n 2 n=1

However,  was arbitrary, and could have been as small as I liked. Therefore, the infimum of the possible lengths of my covers is 0, and so |E| = 0.  As we shall see later, there are also uncountable sets with measure zero. 6

A precise classification of the space of Riemann integrable functions With this notion of Lebesgue measure zero in hand, we can finally tell when a function is Riemann integrable. (Warning: The following proof, while essential, can get a little complicated. If you have questions, post them on the class page rather than smashing your head against the wall). Theorem: A bounded function f is Riemann integrable on [a, b] if and only if its set of discontinuities on [a, b], denoted E, has Lebesgue measure zero Proof: To see that Riemann integrability implies |E| = 0, we shall construct a proof by contrapositive. Suppose that |E| > 0. We wish to show that f ∈ / R([a, b]). To do that, let us define the oscillation of f at a point x ωδ (x) = sup {|f (y) − f (z) : y, z ∈ (x − δ, x + δ)} This tells us precisely how big the discontinuity at x is. If ωδ (x) is large, we have a large discontinuity, and if it is small, we have a minor one. (Show that that if f is continuous at x, the oscillation goes to zero as δ → 0, and visa versa). Now, the problem with our set of discontinuities E is that it may contain a large number of minor discontinuities getting smaller and smaller, whereas we want to use the large ones in our proof. Therefore, we define sets  x ∈ E : ωδ (x) >

En =

1 n

 ∀δ > 0

The set En consists precisely of those points where the oscillation is greater than

E=

∞ [

1 n.

Observe that

En

n=1

Now, if every En had Lebesgue measure zero, so would E (Why?). Thus there exists some natural number N such that |EN | = α > 0. This is very useful, for now we are working with a set where the discontinuities can’t get too small. Now, let P = {a = x0 < x1 < · · · < xk = b} be a partition. We shall use the classical definition of the Riemann integral, where the difference between the upper and lower sums is written k h X

i sup(xi−1 ,xi ) f − inf(xi−1 ,xi ) f (xi − xi−1 )

i=1

This is greater than or equal to X

h

i sup(xi−1 ,xi ) f − inf(xi−1 ,xi ) f (xi − xi−1 )

(xi−1 ,xi )∩EN 6=∅

which is the summation only over those intervals that intersect EN . This in turn greater than or equal to X (xi−1 ,xi )∩EN 6=∅

ωδ (x)(xi − xi−1 ) ≥

1 N

X (xi−1 ,xi )∩EN 6=∅

(xi − xi−1 ) ≥

α >0 N

Since the upper and lower sums will always be some positive distance apart, our function is not Riemann integrable. 7

To prove the converse, suppose that the set of discontinuities E has Lebesgue measure zero. Then |En | = 0 for all n. As an exercise, prove that each En is compact (hint: use the Heine-Borel Theorem). Since En is compact, and has Lebesgue measure zero, it can be covered with a finite collection of intervals, {(ak , bk )} such that X (bk − ak ) <  Let R = [a, b] \

[

(ak , bk )

be the remainder of [a, b] that is not covered. It is easy to see that R is the union of a finite collection of closed sets. By compactness, there exists a δ ∗ such that 1 on R n We can chop up the closed intervals composing R into finitely many subintervals of length less than δ ∗ . Denote these subintervals {(ck , dk )}. Now, let P be a partition whose elements are the ak , bk , ck and dk . With this partition, the difference between the upper and lower Riemann sums is ωδ∗ (x) ≤

k h X

i sup(xi−1 ,xi ) f − inf(xi−1 ,xi ) f (xi − xi−1 )

i=1

Now, let us split this sum in two. The first part will sum over the intervals that do not contain any element of En , and the second part will sum precisely on those that do. =

Xh

i i Xh sup(ak ,bk ) f − inf(ak ,bk ) f (bk − ak ) + sup[ck ,dk ] f − inf[ck ,dk ] f (dk − ck )

Here is where all our hard work pays off. Although the sum of the intervals in the second summation takes up almost all of [a, b], the function itself cannot oscillate more than n1 . In the first sum, although the oscillation may be as large as 2M , the sum of the intervals is very small. Thus b−a n Since  was arbitrary, and we can let n → ∞, we see that the upper and lower Riemann sums can be made equal, and so the Riemann integral exists. . ≤ 2M  +

Note: A property is said to hold almost everywhere if the measure of the set where it fails is zero. As such, one can say that “a function is Riemann integrable if and only if it is continuous almost everywhere”. Exercises: (1) Consider the following variant of the Dirichlet function. Define f (x) on [0, 1] by setting f (x) = 0 when x is irrational. When x is a rational, of reduced form pq , let f (x) = 1q . What is the set of discontinuities of this function? Is it Riemann integrable? (2) Go on Wikipedia and look up the construction of the Cantor set. (a) What is the cardinality of the Cantor set? (b) Let f be a function whose discontinuities lie precisely on the Cantor set. Is f Riemann integrable?

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Measure Zero, Continuity, and the Devil’s Staircase In this mini-lesson, we’ll explicitly construct the Cantor set and the Cantor-Lebesgue function (also known as the Devil’s staircase). If you have completed the homework for Lesson II, you may already know that the Cantor set is interesting because it is an uncountable set of measure zero. For this reason, it is excellent hunting ground for otherwise rare counterexamples. We’ll make use of it in the future, so keep it in mind. The Cantor Set We start with the unit interval [0, 1], and then remove the middle third. This gives us C1 = [0, 1/3] ∪ [2/3, 1] Repeating this process, we remove the middle third from each of these remaining intervals. Then C2 = [0, 1/9] ∪ [2/9, 1/3] ∪ [2/3, 7/9] ∪ [8/9, 1] At each step, we get Ck , the union of 2k intervals of length 3−k . The Cantor Set itself is the intersection of all of all of these Ck , C=

∞ \

Ck

k=1

For the visually inclined, here’s a graphical demonstration of what’s happening:

The Cantor-Lebesgue Function Let Ok = [0, 1] \ Ck be the 2k − 1 open intervals removed at the k th step, and say that O = C C . Define a function ϕ on Ok as follows. Set it equal to 1/2k on the first interval, 2/2k on the second interval, up until (2k − 1)/2k on the 2k − 1 interval. Thus, on O1 , which was just (1/3, 2/3), ϕ is equal to 1/2. Moving up to O2 , we see that   1/4 2/4 ϕ(x) =  3/4

x ∈ (1/9, 2/9) x ∈ (3/9, 6/9) x ∈ (7/9, 8/9)

which coincides perfectly with how we defined it on O1 , as 2/4 = 1/2 (Why?) To extend this function to the Cantor set, we say  0 ϕ(x) = sup{ϕ(t)|t ∈ O ∩ [0, x)}

x∈0 x ∈ C \ {0}

This says: to find out what value ϕ takes on a point x in the Cantor Set, look at the largest value it takes on O before my point. 9

The function we have created is somewhat of a monstrosity, and it will take a little work to prove that it actually behaves rather nicely. It is known as the Cantor function, the Cantor-Lebesgue function, or the Devil’s Staircase, with the last name referring to its pathological graph.

The Strange and Wonderful Properties of the Cantor-Lebesgue Function (1) ϕ is increasing. Prove it. (2) ϕ is constant almost everywhere. Prove it. (3) ϕ is continuous! This is a little harder to prove, but take a shot. (It’s also incredibly counterintuitive, which happens a lot when playing with the Cantor Set) (4) Functions whose derivative is zero almost everywhere are known as singular functions. Show that ϕ is singular. (5) Show that ϕ maps C onto almost all of [0, 1], and O onto the dyadic rationals in [0, 1]. (6) The length of the graph of ϕ on [0, 1] is 2. Prove it. Generalize your technique to all singular functions.

Exercises 1. Use the Cantor-Lebesgue function to construct a continuous, strictly increasing function ψ that maps C onto a set of positive measure. 2. Show that there is a continuous, strictly increasing function that on [0, 1] that maps a set of positive measure onto a set of Lebesgue measure zero. 3. Let f be a Lipschitz function on [a, b]. Show that f maps a set of Lebesgue measure zero onto a set of Lebesgue measure zero. 4. Show that the Cantor-Lebesgue function is not Lipschitz. (It isn’t even Absolutely Continuous, if you’re interested in looking that up on Wikipedia. How is that related to its being singular?) 10

Measure Theory and Lebesgue Integration: Lesson III

Giuseppe Vitali (1875-1932) Lesson 3: Lebesgue Outer Measure. Measurable sets, Non-Measurable sets, and the Axiom of Choice.

Lebesgue Outer Measure An outer measure or exterior measure is a function µ∗ : 2X → [0, ∞] This function takes subsets of my set X and assigns them a size or measure. The range of the function is [0, ∞], because we’d like to think of a set as having positive size. Mathematicians, however, in their unabashed eagerness to do damage to human intuition, have invented the notion of signed measures, which can take negative values. We’ll try to avoid these latter measures for now. Furthermore, an outer measure has the following three properties. (1) µ∗ (∅) = 0 (2) If A ⊂ B, then µ∗ (A) ≤ µ∗ (B). This is called monotonicity. (3) (Countable Subadditivity) Let Aj be a countable collection of sets that may overlap. Then   ∞ ∞ [ X µ∗  Aj  ≤ µ∗ (Aj ) j=1

j=1

I’d like to imagine that if we had developed measure theory independently, we’d also have demanded these three properties, if only because they conform so nicely to our natural geometric intuition. How big is nothing? Not very big at all, I’d imagine. As King Lear so eloquently put it, “nothing can come of nothing,” which accounts for property (1). With regards to the second property, if I can fit one box inside another, I would hope that the outer box would be bigger than the inner one, at least in the context of classical mechanics. Lastly, and with complete respect to Aristotle, the whole should not be any larger than the sum of its parts, especially if those parts can overlap.

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The question, therefore, is not why mathematicians prescribed these three properties, but why they didn’t propose a fourth. Namely, (4) (Countable Additivity) If Ej is a countable collection of pairwise disjoint sets, then   ∞ ∞ [ X ∗  µ Ej = µ∗ (Ej ) j=1

j=1

The answer to this question is rather surprising, but to approach it we must first familiarize ourselves with the Axiom of Choice. The Axiom of Choice Zermelo-Fraenkel set theory, often abbreviated ZF, is a collection of eight axioms upon which much of the rigorous foundation of mathematics is built. Roughly speaking, these axioms tell you what type of sets you are allowed to construct, and were designed by Ernest Zermelo and Abraham Fraenkel at the beginning of the twentieth century, in response to some worrying paradoxes that arose when working haphazardly with sets. The most notable example of this is Russell’s paradox, which I recommend you look up on Wikipedia. In 1904, Ernest Zermelo proposed a ninth axiom, the Axiom of Choice (or AC). Although this axiom was initially controversial, it has been (begrudgingly) accepted, if only because it is crucial to the proof of a number of number of important theorems, such as Tychonoff ’s theorem (for those of you who have taken topology), and equivalent to an ever-expanding list of statements without which math would be a rather sorry subject indeed. Coupled with this axiom of choice, ZF is called ZFC. So what exactly does AC state? Let Si be an arbitrarily large family of non-empty sets. AC guarantees the existence of a collection of points {xi }, with xi ∈ Si . Which is to say, that AC “chooses” an element from each set, and then collects all these points together. Note that the this is a nonconstructive process, as we cannot choose what points to pick, nor can we tell what our collection will look like. To clarify this notion, we turn to a quote by Bertrand Russell: “The Axiom of Choice is necessary to select a set from an infinite number of socks, but not an infinite number of shoes.” The observation here is that AC is not necessary when I have a well-defined algorithm for selecting my points. So, if I specify, “let L be the set of all left shoes”, then I would have selected a point from each set without the axiom of choice. However, since left and right socks are presumably identical, I need AC to make a choice for me. Although this axiom seems like good common sense, it is equivalent to a number of axioms, such as Zorn’s Lemma and The Well-Ordering Principle, that are anything but. As the mathematician Jerry Bona jokingly put it, “The Axiom of Choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn’s lemma?”

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A Word on Equivalence Relationships An equivalence relationship is a relation on a set X that provides a generalized notion of “equality.” When defining an equivalence relationship on a set, I get to specify the requirement for elements to be deemed equivalent. So, for example, I could define an equivalence relationship ∼ on the integers Z as follows: n ∼ m (read: n is equivalent to m) if they are both odd or both even. Thus 1 ∼ 3 and 2 ∼ 4, but 3 is not equivalent to 4. Furthermore, I require an equivalence relationship to have the following three properties: 1. (Reflexive) For all x ∈ X, x ∼ x 2. (Symmetric) If x ∼ y, then y ∼ x 3. (Transitive) If x ∼ y, and y ∼ z, then x ∼ z As the reader may recall from High School Geometry, triangle congruency is an equivalence relationship. The relation, “x is strictly greater than y”, however, is not an equivalence relationship, as it is not reflexive or symmetric. The equivalence relationship I defined on Z gave me two disjoint equivalence classes: odd and even integers. With triangles, the number of equivalence classes is infinite. Vitali’s Theorem and Non-Measurable Sets We are now ready to solve the question I proposed earlier: why did we not include countable additivity in the definition of an outer measure? The solution lies in showing that countable additivity is not compatible with two other properties that are crucial to Lebesgue outer measure (as defined last time): Vitali’s Theorem: There does not exist a function µ∗ : 2R → [0, ∞] such that (1) µ∗ (R) 6= 0,

µ∗ ([0, 1]) 6= ∞

(2) (Translation invariance: shifting a set by some constant does not affect the measure) µ∗ (E + c) = µ∗ (E) (3) µ∗ is countably additive. Proof: Suppose that such a function exists, and define an equivalence relationship on [0, 1] as follows: x∼y

if x − y ∈ Q

This says that x is equivalent to y precisely when the difference between them is rational. Clearly, this equivalence relation partitions [0, 1] into an infinite number of equivalence classes (show that they are disjoint). What’s more, since every equivalence class has only countably many members, and the cardinality of [0, 1] is uncountable, there must be an uncountable number of such equivalence classes. Using the axiom of choice, I produce a set E that contains one element from each equivalence class.

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Note that E can be thought of as generating [0, 1]. Take any element in x ∈ E, and consider {x + q : q ∈ Q and x + q ∈ [0, 1]} This gives the entire equivalence class of x. Repeating this procedure with every element of E, I can reclaim all of my equivalence classes, and hence all of [0, 1]. That being said, let Q∗ = Q ∩ [0, 1], and write [ [0, 1] ⊂ (E + q) q∈

Q

So, what is the measure of E? Case 1: µ∗ (E) = 0. By translation invariance, µ∗ (E + c) = 0. X X µ∗ ([0, 1]) ≤ µ∗ (E + q) = 0=0 q∈

Q

Q

q∈

∗

By translation invariance, µ ([1, 2]) = 0. Since [ R= [n, n + 1]

Z

n∈

countable subadditivity would imply µ∗ (R) ≤

X n∈

Z

µ∗ ([n, n + 1]) =

X n∈

Z

0=0

a contradiction! Case 2: µ∗ (E) = α > 0. Observe that [ q∈

Q∗

(E + q) ⊂ [0, 2]

By translation invariance, µ∗ (E + c) = α > 0. X X µ∗ ([0, 2]) ≥ µ∗ (E + q) = α=∞ q∈

Q∗

Q∗

q∈

If µ∗ ([0, 1]) was some some finite quantity, then µ∗ ([0, 2]) would be twice that finite quantity. Since µ ([0, 2]) = ∞, we must also have µ∗ ([0, 1]) = ∞, which is a contradiction!  ∗

In summary, we have demonstrated, using the axiom of choice, the existence of Non-Measurable sets: sets that cannot be measured in the presence of countable additivity without leading to absurd contradictions. In 1970, Robert Solovay proved that the axiom of choice is essential in constructing non-measurable sets (i.e. they don’t exist in ZF), and thus those of you who do not approve of AC need not worry about the measurability of the sets you are playing with. You also need not worry about getting very far in math. Although Solovay’s result may seem too abstract to be useful, it provides an excellent heuristic when working with Lebesgue measure. If, in the course of proofs or theorems, you come across a family of sets that are defined without AC, you should probably try proving that they are measurable, rather than looking for a counterexample. You won’t find one.

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Measurable Sets Inequalities may be the bread and butter of analysis, but it would be an exercise in futility without equalities. To that end, we need to find a subcollection of 2R where countable additivity works. We shall denote this collection M, and call it “the Measurable Sets.” One says that is a set E is measurable, or E ∈ M, if, for any set A ∈ 2R , |A| = |A ∩ E| + |A ∩ E C | This specifies that the measurable sets are precisely those that “slice” other sets neatly. Take a while to think about this condition, as it will be the central focus in our next lesson.

Exercises: 1) Show that Lebesgue outer measure (as we defined it in Lesson II) is an honest-to-goodness outer measure, possessing the three properties outlined above. 2) Show that every set of Lebesgue outer measure zero is, in fact, measurable. This justifies our use of the term ”Lebesgue measure zero.” 3) Show that every open set is measurable.

15

Measure Theory and Lebesgue Integration: Lesson IV

“Whatever the progress of human knowledge, there will always be room for ignorance, hence for chance and probability.” ´ Emile Borel (1871-1956) Lesson 4: Measurable Sets, Borel Sets and Measure Spaces In the previous lesson, we saw that Lebesgue outer measure | | on 2R possesses the three properties necessary for being an outer measure, but lacked the countable additivity we require of a full-fledged measure. Not wishing to abandon countable additivity or the axiom of choice, we decided to exchange 2R with another collection of sets: M. As we defined it, E ∈ M if |A| = |A ∩ E| + |A ∩ E C |

for any set A ∈ 2R . This specifies that the measurable sets are precisely those that “slice” other sets neatly. At first, it may seem that this condition is somewhat arbitrary, and unrelated to countable additivity. It will take some work to show that M is precisely the collection of sets we want to work with. Algebras and σ-Algebras of Subsets If we are to develop a robust and useful theory of integration, we need our underlying measuretheoretic machinery to be as comprehensive as possible. It won’t do, for example, for our integral to work on [0, 1] but collapse on [2, 17]. In other words, our collection M should be closed under operations such as unions, complements, etc. To make precise this notion, we introduce the idea of an algebra of subsets. An algebra of subsets of X is a subcollection A of 2X such that: (1) X ∈ A. (2) If E ∈ A, then E c ∈ A. (3) If E1 , · · · , En ∈ A, then

n [

Ej ∈ A.

j=1

16

Combining the second property with the first tells us that ∅ ∈ A, and the second and third properties together prove that A is closed under finite intersections, as, by De-Morgan’s Law,  c n n [ \  Ejc  = Ej j=1

j=1

Lastly, if our collection A is closed under countable (and not just finite) unions, we say that A forms a σ-algebra of subsets of X. Proving that M is a σ-algebra of subsets of

R

It’s rather easy to prove that R ∈ M, and that M is closed under taking complements, so I’ll leave that to the reader as an exercise. Theorem: M is closed under finite unions Proof: By induction, it will suffice to show that the union of two measurable sets is measurable. Let E1 , E2 ∈ M, and let A ∈ 2R . By countable subadditivity, |A| ≤ |A ∩ (E1 ∪ E2 )| + |A ∩ (E1 ∪ E2 )c | Applying countable subadditivity again, |A ∩ (E1 ∪ E2 )| + |A ∩ (E1 ∪ E2 )c | ≤ |A ∩ E1 ∩ E2 | + |A ∩ E1 ∩ E2c | + |A ∩ E1c ∩ E2 | + |A ∩ E1c ∩ E2c | Conceptually, this sum contains four terms because a point in A can be covered by E1 , E2 , both, or neither. Now, by the measurability of E2 , we know that |A ∩ E1 ∩ E2 | + |A ∩ E1 ∩ E2c | = |A ∩ E1 | |A ∩ E1c ∩ E2 | + |A ∩ E1c ∩ E2c | = |A ∩ E1c | Thus, by the measurability of E1 , |A ∩ E1 ∩ E2 | + |A ∩ E1 ∩ E2c | + |A ∩ E1c ∩ E2 | + |A ∩ E1c ∩ E2c | = |A ∩ E1 | + |A ∩ E1c | = |A| This demonstrates that |A| ≤ |A ∩ (E1 ∪ E2 )| + |A ∩ (E1 ∪ E2 )c | ≤ |A| so that equality holds, and we may conclude that M is an algebra of subsets.  Theorem: Lebesgue outer measure, restricted to M, is finitely additive Proof: Let E1 , · · · , En ∈ M be pairwise disjoint, and let A=

n [

Ej

j=1

Then |A| = |A ∩ E1 | + |A ∩ E1c | = |A ∩ E1 | + |A ∩ E2 | + |A ∩ E2c | = · · · =

n X j=1

17

|Ej | 

Theorem: M is closed under countable unions

R Proof: Let {En }∞ n=1 be a countable sequence of measurable sets, and let A ∈ 2 . Assume that ∞ {En }n=1 is pairwise disjoint, otherwise we can replace it with   n−1 [ En∗ = En \  Ej  j=1 ∗

which are pairwise disjoint, noting that ∪E = ∪E. By countable subadditivity, ∞ [ |A| ≤ A ∩ En + A ∩

∞ [ n=1

n=1

!c En

If |A| = ∞, the reverse inequality is trivially true, and equality holds. Otherwise, we can bound the two terms in right hand side of the above equation. ∞ ∞ ∞ X [ [ ≤ |(A ∩ En )| (A ∩ E ) A ∩ E = n n n=1

A ∩

n=1

n=1

!c ∞ [ En ≤ A ∩

N [

n=1

n=1

!c En

∀N ∈ N

Combining these inequalities, we find that !c (N ∞ ∞ [ [ X En + A ∩ En ≤ lim |(A ∩ En )| + A ∩ A ∩ N →∞ n=1

n=1

n=1

N [ n=1

!c ) En

Observe that the interior of the limit contains only finite intersections and sums. Since we proved finite additivity, !c N N X [ |(A ∩ En )| + A ∩ En = |A| n=1

n=1

Substituting, ∞ [ En + A ∩ A ∩ n=1

∞ [ n=1

!c En ≤ lim |A| = |A| N →∞

This reverse inequality completes the proof. 

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Theorem: Lebesgue outer measure, restricted to M, is countably additive Proof: Let {En }∞ n=1 be a countable sequence of pairwise disjoint measurable sets. Since M is a σ-algebra, the union of these sets is measurable. By countable subadditivity, ∞ ∞ [ X En ≤ |En | n=1

n=1

By monotonicity of measure and finite additivity, ∞ n n X [ [ En = |En | En ≥ n=1

n=1

n=1

Letting n → ∞, we conclude that ∞ ∞ [ X En = |En | n=1



n=1

Although the proof above did not directly require that M be a σ-algebra, it would be quite useless without that fact. This is because we constructed M to avoid the possibility of non-measurable sets. If the union of the collection of measurable sets were not measurable, we would avoid assigning it a measure, much as we avoid assigning the Vitali set a measure. It is only when we are assured that our set can be measured that the precise value of its measure is of consequence. By this point, we should all be thoroughly convinced that M is the ideal collection of subsets to work with. We will now show that M also contains most of the sets we know of. One of the exercises from Lesson 4 was to show that the open sets are measurable. A consequence of this is that M contains all Borel sets. Borel Sets To create the Borel sets B, we start with all open sets, and take all possible sequences of complements (giving us all closed sets), countable unions and countable intersections. We already know that the arbitrary union of open sets is open, but the Borel sets also contain the countable intersection of open sets, known as Gδ sets. The letter “G” is taken from the German word “Gebiet” meaning ”area”, and the δ stands for the word “Durchschnitt”, meaning intersection. Similarly, I can take the countable union of closed sets to get Fσ . The Letter “F” stands for the French word “ferm´e”, meaning “closed,” and the σ is short for “somme”, meaning union. But why stop there? Taking the countable union of Gδ sets, we get Gδσ sets, etc. The list is literally endless, and we could go on and on defining Fσδσδσ sets. However, we can also make the quick assertion that the Borel sets are the smallest σ-algebra containing the open sets (an idea that should be familiar to those who have taken topology). By extension, M is also a σ-algebra of sets containing the open sets, and hence B⊂M In fact, B is a proper subset of M, and we’ll prove this next time. Measure Spaces 19

A measure space (X, A, µ) is a triple consisting of a set X, a σ-algebra of subsets A, and a countably additive measure µ. By the prior theorems, we have established that (R, M, | |) is a measure space. Exercises: 1) Prove that the following are equivalent: a) E ∈ M b) ∀ > 0, there exists an open set U such that E ⊂ U and |U \ E| <  c) There exists a Gδ set S such that E ⊂ S and |S \ E| = 0 2) Prove that the homeomorphic image of a Borel set is Borel. 3) Show that if E has positive outer measure, then there is a bounded subset of E that has positive outer measure. (Taken from Royden’s Real Analysis) 4) Show that if E has finite measure, and  > 0, then E is the disjoint union of a finite number of measurable sets, each of which has measure at most . (Taken from Royden’s Real Analysis)

20

Measure Theory and Lebesgue Integration: Lesson V

“In my opinion, a mathematician, in so far as he is a mathematician, need not preoccupy himself with philosophy – an opinion, moreover, which has been expressed by many philosophers..” Henri Lebesgue (1875 - 1941) Lesson 5: Measurable functions, Simple functions, and the four-step construction of the Lebesgue integral.

Measurable Functions In this lesson, we will use the measure space (R, M, | |) to finally construct the Lebesgue Integral. There is, however, one more thing to attend to. We need to make sure that the functions we will integrate cannot allow non-measurable sets to enter the equation. Before we attend to that, let us define the characteristic (or indicator) function χE , also written

1E . This function is 1 on E and 0 elsewhere, and hence “indicates” the set E. Throughout the rest of this course, we will use make judicious use of characteristic functions in proofs and definitions, as they allow us great flexibility in manipulating functions. Definition: A function f , defined on a measurable set E, is called measurable if any of the following four conditions hold (show that they are equivalent): 1) For each c ∈ R, the set {x ∈ E | f (x) > c} is measurable. 2) For each c ∈ R, the set {x ∈ E | f (x) ≥ c} is measurable. 3) For each c ∈ R, the set {x ∈ E | f (x) < c} is measurable. 4) For each c ∈ R, the set {x ∈ E | f (x) ≤ c} is measurable. Each of these properties imply that f −1 (c) ∈ M for any c ∈ R. Proposition 1: A function f is measurable on E if and only if f −1 (O) is measurable for any open set O. 21

Proof: If f −1 (O) is measurable for any open set O, then condition (1) of the definition is satisfied, as (c, ∞) is an open set for any c ∈ R. Conversely, suppose f is measurable, and let O be an open set. By a well-known result, O is the union of a countable collection of open intervals {Ik } = (ak , bk ). Each of these intervals (ak , bk ) is itself the intersection of two intervals Ak and Bk , where Ak = (ak , ∞)

Bk = (−∞, bk ) Thus we can write " f

−1

(O) = f

−1

∞ [

# Ak ∩ B k =

k=1 −1

∞ [  −1  f (Ak ) ∩ f −1 (Bk ) k=1

−1

Since f is measurable, f (Ak ) and f (Bk ) are measurable, and because the measurable sets form a σ-algebra, we may conclude that f −1 (O) is measurable.  This proposition immediately implies that continuous functions are measurable, as we would have hoped. Furthermore, Proposition 2: A monotonic function defined on an interval is measurable. Proposition 3: If f is measurable on E, and f = g almost everywhere, then g is measurable on E. Proof: The proof of these propositions will be left as exercises to the reader. Simple Functions and Simple Approximations In the first lesson, we defined a step function to be one that assumes a finite number on values on a collection of intervals. Generalizing that idea, we define a simple function ψ to be one that assumes a finite number of values on a collection of measurable sets. Such simple functions are always measurable, and we will use them to construct the Lebesgue integral, much like we used step functions to build the Riemann integral. To see the power of working with simple functions, consider the Dirichlet function. Because the Dirichlet function was defined on a complicated set like the irrationals, one could not approximate it with step functions, and hence it was not Riemann integrable. However, because the Dirichlet function is a simple function (why?), it fits naturally into the framework of the Lebesgue integral. To strengthen the connection between measurable functions and simple functions, we will prove a key result known as the Simple Approximation Theorem. Simple Approximation Lemma: Let f be a bounded, measurable, real-valued function defined on a measurable set E. For any  > 0, there are simple functions ψ and ϕ defined on E such that ϕ ≤ f ≤ ψ

0 ≤ ψ − ϕ < 

Proof: Since f is bounded, its image is contained in some bounded, open interval (c, d). Let P = {c = x0 < x1 < · · · < xn = d}

22

be a partition of this open interval such that xk − xk−1 < . Define Ik = [xk−1 , xk ), and let Ek = f −1 (Ik ). Since f is a measurable function, Ek is a measurable set. Define the simple functions ϕ and ψ to be ϕ =

n X

yk−1 χEk

ϕ =

k=1

n X

yk χEk

k=1

By construction ϕ ≤ f ≤ ψ and 0 ≤ ψ − ϕ < , and so the proof is complete.  Roughly speaking, we took the bounded range of our function f and chopped it into a collection of intervals less than  in length. Since f was measurable, we found the measurable sets on which these values were attained, and defined ϕ and ψ to assume the smallest and largest values on that set, respectively. As a result, these simple functions are never more than  away from f .

Simple Approximation Theorem: An extended real-valued function f on a measurable set E is measurable if and only if there exists a sequence of simple functions ψn converging pointwise to f on E with the property that |ψn | ≤ |f | Proof: Suppose that there exists a sequence of simple functions ψn converging to f on E. We wish to show that f is measurable, so that for any c ∈ R, {x ∈ E : f (x) < c} ∈ M Now, if f (x) < c, and limn→∞ fn (x) = f (x), then there must exist natural numbers k and n such that 1 ∀j ≥ k n This means that at some point in my sequence of functions, they need to send x to a value less than c − n1 . Fixing a value of n, the set of all points that are eventually mapped to values less than c − n1 is fj (x) < c −

∞ \

{x ∈ E | fj (x) < c −

j=k

1 } n

If we take into account all possible values n, we obtain all the points that are eventually mapped to a value less than c. Hence {x ∈ E : f (x) < c} =

∞ \ ∞ [

{x ∈ E | fj (x) < c −

k,n=1 j=k

1 } n

Since the fj are measurable, and the measurable sets form a σ-algebra, the sets above are measurable, and so f is measurable. Conversely, suppose that f is a measurable function. Since f may not be bounded, we cannot directly appeal to the Simple Approximation Lemma. Instead, we will first “pretend” that f is bounded, by restricting it a subset En of E, such that En = {x ∈ E | f (x) ≤ n}

23

Now, use the Simple Approximation Lemma to conjure up sequences of simple functions ϕn and ψn defined on En such that 0 ≤ ϕn ≤ f ≤ ψn and 0 ≤ ψn − ϕn
n. The idea is as follows: since we cannot approximate unbounded functions with simple functions, we artificially bound f with a ceiling of height n. As n increases, the ceiling gets higher, and the simple functions ϕn get closer to f . If f (x) = ∞, it will always be above our ceiling, so we will set ϕn = n for all n ∈ ϕn (x) → ∞, and we have pointwise convergence.

N. As n → ∞,

If f (x) is finite, it is less than some natural number N . Then 0 ≤ f (x) − ϕn (x)
0, and is equal to zero otherwise. Similarly, f − (x) = −f (x) when f (x) < 0, and is equal to zero otherwise. This “decomposes” f into two nonnegative, measurable functions, such that f = f+ − f−

|f | = f + + f −

If |f | (a nonnegative, measurable function) is integrable, we say that f is integrable, and we write f ∈ L1 (E). The integral of f is Z Z Z f= f+ − f− E

E

E

The reason for first requiring that |f | be integrable will be made clear below.

25

One Minor Complication Since the Riemann integral could only integrate over finite intervals, the positive and negative values of the function would “cancel” each other out. Consider the following periodic function f ,

With the Riemann or Lebesgue Integral, Z lim n→∞

nπ

f = lim 0 = 0 n→∞

−nπ

Suppose, however, that we wish to integrate directly over R using the Lebesgue integral. Z Z f+ = f− = ∞

R

R

As a result, Z f =∞−∞

R

which is not well-defined, and so f ∈ / L1 (R)! This is why we first require that |f | be integrable, + which is equivalent to requiring that f , f − ∈ L1 (R). In other words, the positive and negative values of the Lebesgue integral are calculated first and added together later. In order to allow the positive and negative values to cancel each other out, we would have to take the limit of a sequence of integrals, much like defining an improper Riemann integral. Note: A function is said to be locally integrable, written f ∈ L1loc (E), if every point has an open neighborhood on which f is integrable. Our periodic function above, although not integrable over R, is locally integrable. Exercises: (1) Prove Proposition 2. (2) Prove Proposition 3. (3) Let f ≤ g, and suppose that f, g ∈ L1 (E). Show that Z Z Z αf + βg = α f +β g E

E

Z

E

Z f≤

E

g E

(a) for simple functions. (b) for bounded, measurable functions of finite support. (c) for nonnegative, measurable functions. (d) for general Lebesgue-integrable functions.

26

(4) Let f be integrable over E, and let A and B are disjoint measurable subsets of E. Use the result of problem (3), together with the χ function, to show that Z Z Z f= f+ f A∪B

A

B

1

(5) Show that if f, g ∈ L (E), and f = g almost everywhere, then Z Z f= g E

E

27

Measure Theory and Lebesgue Integration: Lesson VI

Besides for being good-looking and French, Pierre Fatou developed key results in Integration Theory and Complex Analytic Systems. In fact, Fatou was the first mathematician to define the Mandelbrot set! Pierre Fatou (1878 - 1929) Lesson 6: Littlewood’s Three Principles. The Bounded and Uniform Convergence Theorems. Fatou’s Lemma and the Dominated and Monotone Convergence Theorems.

The theorems we are about to prove are a pretty big deal, and will make the Lebesgue integral a god-damned pleasure to work with. Trust me, once you’ve used the Lebesgue integral, there’s no going back.

28

Littlewood’s Three Principles In his 1944 text, Lectures on the Theory of Functions, J.E. Littlewood outlined three theorems that give great insight into the essentials of measure theory. These three principles give one an intuitive way of thinking about measurable sets and functions, something that will prove invaluable as we make the transition to advanced analysis. Due to the subtle and nontrivial nature of these proofs, I will refrain from demonstrating them here, and will rather present only the results. Principle One: Every measurable set of finite outer measure is almost the finite, disjoint union of open intervals: Let E ∈ M such that m∗ S (E) < ∞. For each  > 0, there exists a finite, disjoint collection of open n intervals {Ik }nk=1 , with O = k=1 Ik , such that m∗ (E4O) = m∗ (E ∼ O) + m∗ (O ∼ E) <  Note: The symbol 4 above denotes the symmetric difference of two sets. A4B consists of those points that are either in A or B, but not both.

Principle Two (Egoroff ’s Theorem): Every pointwise-convergent sequence of measurable functions is nearly uniformly convergent: Let E have finite measure, and let fn be a sequence of measurable functions converging to a real-valued function f on E. For every  > 0, there exists a closed set F contained in E such that m(E ∼ F ) < , and fn → f uniformly on F Principle Three (Lusin’s Theorem): Every measurable function is nearly continuous: Let f be a real-valued measurable function on E. For any  > 0, there exists a continuous function g on R and a closed set F ⊆ E such that m(E ∼ F ) <  and f = g on F .

29

The following two convergence theorems are concerned with bounded functions of finite support. Uniform Convergence Theorem Let fn be a sequence of bounded, measurable functions on a set E of finite measure. If fn → f uniformly on E, then Z Z lim fn = f n→∞

E

E

We refer to this as the passage of the limit under the integral sign. Proof: Let  > 0. Since our sequence converges uniformly to f , there must exist an N ∈ that |fn − f |
n} = 0

n→∞

Since f χ{x∈E|f (x)>n} ≤ f , we apply the MCT to conclude that Z Z lim f χ{x∈E|f (x)>n} = 0=0 n→∞

E

E

And thus for any  > 0, there exists a natural number N such that Z f χ{x∈E|f (x)>N } <   E

32

Dominated Convergence Theorem (DCT) Let fn be a sequence of measurable functions on E. Suppose that there exists a function g that is integrable on E so that |fn | ≤ g. Such a function is said to dominate our sequence. If fn → f almost everywhere on E, then f is integrable on E, and Z Z lim fn = f n→∞

E

E

Proof: g − fn is positive, so by Fatou’s Lemma, Z  Z Z Z Z Z g − fn ≤ lim inf (g − fn ) = lim inf g− fn = g − lim sup fn E

E

E

E

E

E

Note that we switched from the “lim inf” to the “lim sup”, because the smallest value of the integral occurs precisely when the fn are largest. Subtracting

R E

g from both sides, and dividing by −1, we find that Z Z f ≥ lim sup fn E

E

Similarly, Z  Z Z Z Z Z g + fn ≤ lim inf (g + fn ) = lim inf g+ fn = g + lim inf fn E

E

E

E

E

E

so that Z

Z f ≤ lim inf

E

fn E

Combining these inequalities, we see that Z Z Z Z Z lim inf fn ≤ lim sup fn ≤ f ≤ lim inf fn ≤ lim sup fn E

E

E

Z lim inf

E

Z fn =

E

f = lim sup E

Z lim

n→∞

E

Z

Z fn =

E

f E

33

fn E



Here’s a rough outline of of the connection between the convergence theorems we’ve proven in this lesson.

Exercises: 1) Prove the Generalized Dominated Convergence Theorem. The statement is almost identical to that of the DCT, expect we replace g with a sequence of nonnegative measurable functions gn such that |fn | ≤ gn . We also specify that the sequence convergence pointwise to some function g almost everywhere on E, and that Z Z lim gn = g≤∞ n→∞

E

E

2) Let f be a nonnegative measurable function on R. Show that Z n Z lim f= f n→∞

R

−n

(Taken from Royden’s Real Analysis) 3) Let fn be a sequence of integrable functions on E such that fn → f almost everywhere on E, and f is integrable on E. Show that Z |f − fn | → 0 E

if and only if Z |fn | =

lim

n→∞

Z

E

|f | E

Hint: use the generalized DCT. (Taken from Royden’s Real Analysis)

34

Challenge Exercise: A function has compact support if the space on which it is nonzero is compact. We denote the space of continuous functions of compact support C0 (R). Prove that C0 (R) is dense in L1 in the L1 norm, i.e. Z d(f, g) = |f − g| Hint: use Lusin’s Theorem, the Tietze Extension Theorem and a convergence theorem of your choice.

35

Measure Theory and Lebesgue Integration: Lesson VII

Lesson 7: The Riesz-Fischer Theorem: L1 is complete.

When we began this course, we highlighted three concerns with the Riemann integral. The first was the large class of non-integrable functions, the second was the inability to integrate over infinite sets, and the third was the fact that R was not complete. We have shown that the Lebesgue integral can integrate over almost any function we can imagine, and that it can be defined on sets of infinite measure. In the last lesson, we demonstrated a number of useful convergence theorems that make the Lebesgue integral not only more versatile than its Riemann counterpart, but often easier to use. To complete (no pun intended) our program of a finding “a better integral,” we will now show that L1 is complete, and hence a Banach space. L1 Convergence vs. Pointwise Convergence Before we directly attempt the proof of the Riesz-Fischer theorem, we must take note of a potential complication. For a sequence of functions to converge in the L1 norm, it is not necessary that they converge pointwise almost-everywhere. In fact, they may not converge pointwise anywhere! An example of this can be found in the following sequence of functions: fn,k = χ[ k−1 n , 2

k 2n

]

for n ∈ N, k ∈ {1, · · · , 2n }

Here is a depiction of the first six terms in this sequence:

36

Each time we increment n, the length of the interval shrinks, and then as k ranges from 1 to 2n , that interval marches along [0, 1]. Clearly, the size of the integral gets smaller and smaller, so this sequence converges to 0 in the L1 norm. Nevertheless, for any given point in [0, 1], our sequence of intervals will continue to hit that point, no matter how large n and k become. Thus this sequence does not converge pointwise to 0 anywhere! Note, however, that if we fix k = 1, we get a subsequence that converges pointwise to 0 almost everywhere (except at the origin). This will be our line of attack: we will take our Cauchy sequence, extract an almost-everywhere-pointwise-convergent subsequence, and then apply the dominated convergence theorem to that. Theorem (Riesz-Fischer) : Let fn be a cauchy sequence of fuctions in L1 (E). Then fn converges to some function f ∈ L1 (E), and Z

Z

lim

n→∞

fn =

f

E

E

Proof: Since fn is a cauchy sequence, there exists some n() ∈ N such that ∀n ≥ n(), Z |fn − fn−1 | <  E

Now, define a subsequence nk = max (nk−1 , n(1/2k )). We will show that this subsequence converges pointwise almost everywhere. Set g(x) = |fn1 (x)| +

∞ X

|fnk+1 − fnk |

k=1

Taking the integral, Z

Z |fn1 (x)| +

g= E

Z X ∞

E

|fnk+1 − fnk |

E k=1

By the monotone convegence theorem, we can pull the integral into the summation (why?) Z |fn1 (x)| +

= E

∞ Z X

|fnk+1 − fnk |

E

k=1

Using the bounds that we built into our subsequence, Z ≤

|fn1 (x)| + E

∞ X 1