## A Quick Note on Fatou’s Lemma.

### August 16, 2011

Above is a sweet picture of this French mathematician Pierre Fatou.  Of course he had a nice mustache, but a few math things that are kind of neat about him: he was the first person to study the Mandelbrot set (though there is some controversy regarding the word "first" here…) and he also enjoyed iterative and recursive processes before they became really cool, making him a computer science hipster.  Also, he has a lemma in measure theory named after him.  That’s what this post is about.

Lemma (Fatou’s):  Let $\{f_{i}\}_{i}$ be a sequence of non-negative measurable functions on ${\mathbb R}$ with the Lebesgue measure $\mu$,  and let $\displaystyle f(x) = \lim_{n\rightarrow\infty}\inf_{m \geq n} f_{n}(x)$ for all $x\in {\mathbb R}$.  Then, $f$ is measurable and we have the inequality

$\displaystyle \int f\, d\mu \leq \lim_{n\rightarrow\infty}\inf_{m \geq n} \int f_{n}\, d\mu$.

## Nested Sequences of Measurable Sets.

### August 10, 2011

On nearly every practice qualifying exam that I’ve been studying from, the following question (in some guise) comes up:

Question: Let $\{E_{i}\}_{i=1}^{\infty}$ be a sequence of Lebesgue measurable sets in ${\mathbb R}$.  Denote the Lebesgue measure by $\mu$

1. If $E_{n}\subseteq E_{n+1}$ for every $n\in {\mathbb N}$, then is it true that $\displaystyle \mu(\bigcup_{i=1}^{\infty} E_{i}) = \lim_{n\rightarrow\infty}\mu (E_{n})$?  If not, add a criteria to make it true.
2. If $E_{n}\supseteq E_{n+1}$ for every $n\in {\mathbb N}$, then is it true that $\displaystyle \mu(\bigcap_{i=1}^{\infty} E_{i}) = \lim_{n\rightarrow\infty}\mu (E_{n})$? If not, add a criteria to make it true.

I’ll write up the solutions, since they are available elsewhere, but take a little bit of time to think about the second one if you haven’t.  I’ll put the remainder of the post under a cut so that I don’t spoil it right away.

## Brief "Where are we going?" Motivation.

So far we’ve been talking about measure theory, which is nice to study just for the sake of measuring things, but when we talk about functions and measurements, we generally want to know the integral.  The Riemann integral was wonderful because of its simplicity in defining it: you have a big sheet of wavy paper (the total area under a function) and you keep slicing the pieces until they look like rectangles — once they look sufficiently close to a rectangle, just use length-times-width, and add up all the rectangles.  This is one interpretation of the Riemann integral, and, for the most part, it works wonderfully.  But there are many functions where it fails miserably.  This is where Lebesgue takes over.

So what’s different about Lebesgue?  While Riemann cuts apart the domain, Lebesgue says, "Okay, how large is the set of x-values that have the range value $c\in {\mathbb R}$?"  Specifically, we cut up the function into sets which have the same range value, then multiply that range value with the measure of the set.

As an example, take the step function $f$ defined on, say, $[-10,10]$ such that $f(x) = 1$ for $0\leq x < 1$ and $f(x) = 2$ for $1\leq x < 2$ and $f(x) = 0$ elsewhere.  The Lebesgue integral will say, "Where does $f$ take on the value 0?  Well, $[-10,0)\cup(2,10]$ which has measure of 18.  Since the value is 0, we have $18\cdot 0 = 0$, so this contributes nothing to the integral.  Where does $f$ take on the value 1?  On $[0,1)$, which has a measure 1.  So this contributes $1\cdot 1$ to the integral.  Similarly, the function takes on a value 2 onn $[1,2)$, this has measure 1, so this contributes $1\cdot 2$ to the integral.  No other values are taken on, so this integral is 0 + 1 + 2 = 3."

This extremely simple example shows the difference between Riemann and Lebesgue at the most fundamental level.  To do Riemann would be much easier in this case: it’s just adding up the rectangles.

There is a slight hitch at this point. Suppose we naively tried to apply what we’ve just said to a more complex function like $f(x) = x$ defined, say, on $[0,2]$.  We know (because this forms a triangle) that the area should be 2.  At every number between $[0,2]$ in the range, though, the function only takes on the value at a single point which has measure 0.  Thus, every one of these contributes 0 to the integral; and the sum of 0’s will be 0.  So the integral is 0?  That’s not right!

The problem here is that we must define the Lebesgue integral for functions which take on simple functions (which take on a countable number of values) and somehow approximate more general functions.  As we know, a function is measurable if and only if there is a sequence of simple functions which converge uniformly to it; and this will give us the means to integrate general (measurable) functions: we will take the integrals of each element of the sequence, and then define that limit to be the integral of the measurable function.  A number of questions will arise instantly, like, "Does this work?  Does this make sense?  Is this well-defined?", all of which can easily be put down with the elegant slashing of our proof-swords.

So, let’s get down to it.

## The Lebesgue Integral and Simple Functions.

We’ve already gone over simple functions (ones with a countable number of values, the pre-image of each of these being measurable making simple functions, themselves, measurable) so we might as well dive right in.  We will begin by defining, after long last, the Lebesgue Integral for these simple functions.  It will be defined exactly as we have motivated it above.

Definition.  Let $f$ be a measurable simple function on a (measurable) set $A$ taking the values $a_{1}, a_{2}, \dots$.  Let $A_{i}$ be defined as the set of all $x$ such that $f(x) = a_{i}$; that is, $A_{i}$ is the pre-image of $a_{i}$.  Then we define

$\displaystyle \int_{A}f(x)\, d\mu$

to be exactly the sum

$\displaystyle \sum_{i=1}^{\infty} a_{i}\mu(A_{i})$

provided that the sum converges absolutely (that is, the sum converges with $|a_{i}|$ replacing $a_{i}$).

We call this the Lebesgue integral of $f$ over the set $A$.

Note that we have explicitly noted that $A$ must be measurable, though this is not specifically stated in the texts I am using.  I don’t feel I am losing much in assuming this (expect, perhaps, for more pathological examples such as being able to integrate some function which is non-zero only on a measurable subset of the non-measurable set).  Notice that we have also assumed that our sum must converge absolutely, and not just conditionally.

[Note: At this point, it is not entirely clear to me why this should be necessary; in finite cases, we will generally not worry about it, but perhaps there is some reasoning as to why in infinite measured sets, we need this absolute qualification.  Feel free to weigh in here.]

It is reasonable to want to check the well-defined-ness of this, but I will not do this here since it is available where ever Lebesgue integrals are gone over.  I will also not define the integral of $f + g$ or $\alpha f$ for $\alpha\in {\mathbb R}$, because these should be relatively obvious.

TO BE CONTINUED.

## Real Analysis Primer, Part 7: Measurable Functions and Limits.

### July 21, 2011

Limits play an important role in standard calculus: it is the limit that allows us to get “infinitely small” and to be able to take derivatives and integrals. It would be nice, then, to see when our functions behaved nicely when we took limits of them.

In fact, we have a relatively nice theorem regarding sequences of measurable functions and their limits. Why do I pick this theorem to prove for you out in particular, as opposed to the many others above which I simply state? I feel that the proof of this theorem is neither a “follow-your-nose” proof, nor an “ah-ha!” proof that you will “get” once you read it; no, this proof is more of a “how was I supposed to think of that!” proof. What’s more is that the proof is not complicated; in fact, after explaining exactly what is going on (which I will do, following the proof) it will seem nearly obvious to you.

## Motivation.

I’m a bit sad that I don’t have a catchy title for this post, but at least we have a neat topic: measurable functions.  In pure, unadulterated generality, these things (to me, at least) look exactly like continuous functions in topology: the idea is that if you "pull back" a measurable set, you ought to get a measurable set.  Specifically, our "pulling back" in this case is doing to be taking the pre-image of the function (not to be confused with the inverse, which does not always exist; the pre-image, on the other hand, exists and is a set). Does this look familiar?  If I said, maybe, "open" instead of "measurable"?  Yeah.  Yeah.

In this post, we will work with ${\mathbb R}$ with the standard sigma-algebra generated by the open sets of the standard topology on ${\mathbb R}$.  But, that’s still kind of a lot of sets to consider.  Do we have to look at them all, or do we have a basis for this sigma-algebra?

For all you lazy mathematicians, you’re in luck.  In fact, the sets $(-\infty, a)$ for each $a\in {\mathbb R}$ generate the sigma-algebra we’re considering (try to prove this; it’s not that hard!)  Thus, it suffices just to check out the pre-images $f^{-1}((-\infty,a))$ for $a\in {\mathbb R}$ in order to see if $f$ is measurable.

Last, note that we could use the other open or closed semi-infinite intervals, but I like $(-\infty, a)$ the best.

## Real Analysis Primer, Part 5: Measurable Sets are Pretty Close To Stuff We Like.

### June 27, 2011

I was going to write about measurable functions, but after a little discussion about measurable sets with a friend, I decided to put that off for a bit.  The discussion had to do with the nature of measurable sets and how weird they could look.  He notes (paraphrased!):

"I mean, open sets are measurable.  Those are nice.  And closed sets are measurable.  Those are nice."

"Uh-huh."

"But there’s all sorts of sets which aren’t open or closed which are measurable.  I mean, there’s sorts of sets which don’t even look like finite unions of open and closed sets."

"Uh-huh."

"So to say that measurable sets are ‘nice’ in some way is really not an accurate statement.  We can make a measurable set as not-nice as we want it!"

## Mild Introduction.

Last time, we described some facts about Lebesgue measure and showed that despite being a nice measure of sets on the real line there are some sets which cannot be measured by the Lebesgue measure.  Intuitively, this gives us a big contrast with “real world” objects — there are very few things (except for the very, very small and the very, very large, perhaps) which we cannot hold a ruler next to and measure or, at least, approximate.  The previous set we constructed could not even be approximated — or, at least, it is not obvious how to approximate such a beast!

On the other hand, we will show that there are sets which, informally speaking, the Lebesgue measure “doesn’t care” about.  These sets are called sets of measure zero, and they’re quite easy to describe.

Definition.  We say that $A$ is a set of measure zero if $\mu(A) = 0$.

These sets will act like a “zero” element with measures.  If we have some set $B$, then it will turn out that if $A$ has measure zero, we will show that $\mu(B) = \mu(B\cup A)$.  This is more or less what I mean when I say that the Lebesgue measure does not care about these sets.

[Note: from now on, I will use the term “measurable” to mean “Lebesgue measurable” and I will qualify when I don’t mean this.]

## Last time, and some Equivalent Statements.

Last time we talked about the Lebesgue measure in terms of outer and inner measure.  Recall that outer measure of a set is this "almost-measure" that we made up by taking the infimum of sets that looked like disjoint intervals which covered the set we’re measuring.  Recall that we worked in the unit interval $[0,1]$ and recall that $E\subseteq [0,1]$ is an elementary set if it is a countable union of intervals (half-open, open, or closed).  Then we can better write outer measure as

$\displaystyle \mu^{\ast}(A) = \inf_{A\subseteq E}\{\mu^{\ast}(E)\}$

and we also had a corresponding "inner measure" which was another almost-measure, defined in the following way:

$\mu_{\ast}(A) = 1 - \mu^{\ast}([0,1]-A)$

and we pieced these together to make an actual measure, called the Lebesgue measure,  which is defined on every set where the inner and outer measures match up.  In other words, if $\mu^{\ast}(A) = \mu_{\ast}(A)$, then $\mu(A)$ exists and $\mu(A) = \mu^{\ast}(A)$

## Review and Motivation!

Last time we constructed the notion of a measure in a pretty general setting.  What was it?  It was a function from a nice set of subsets (a sigma-algebra) to the non-negative reals union infinity that satisfied a relatively reasonable condition: if we were to union up a bunch of disjoint measurable sets, the measure of the union should be the sum of the measure of the sets.  That’s a pretty general kind of function.  But even the most unobservant of readers wouldn’t be able to help but notice my lack of creativity when talking about measures — they all related to the real line!  The reader should note at this point that there are other kinds of measures out there; in fact, there are measures which are wild and crazy and have nothing to do with the intuitive notion of "length"!  We may be touching on some of these measures in later posts.

The real line is comfortable.  We know a lot about it.  We can picture it.  We can even apply it to things in the real world.  Thus, if we start talking about measures, a good place to start building them would be on the real line.  And if measures are supposed to, somehow, be this generalization of "length" or "size", then what better a way to make a measure on the subsets of the real line than to have our measure give us the total length of the subset!

And this is the general idea behind Lebesgue measure.  So what’s the big deal?  Well, remember when we defined measures before, we needed to define them on a sigma-algebra.  It turns out that this kind of measure doesn’t work so well on the powerset of the reals: indeed, there are subsets of the real line which are not measurable (!) in the way that we just vaguely defined it.  The first time I learned about this, it blew my mind.  And it still does.  How could something so simple go wrong?

Maybe we are expecting too much.  Let’s start modestly.  At the very least, we can talk about intervals — we know what length we’d like to assign to them.

[In this post, I have made liberal use of additional notes, which I will put in-between braces and in italics, just like the sentence you are reading right now!  These additional notes can be skipped with no major loss of understanding the subject, and only provide additional rigor or verify statements which are necessary but break the flow of the post.  As I’ve mentioned before in this blog, my intention is NOT to write the Great American Mathematics text book, but simply give motivation and a general argument as to how we do what we do.  Last, this post, for the curious reader, follows Kolmogorov’s Introduction to Real Analysis text, section 25. My exposition is slightly different, but I don’t think there should be any problems with it.]

## Introduction.

Every so often we do things not because we want to, but because we must.  It is not a secret that I do not like Analysis.  It may be because, as some of my peers suggest, I don’t truly understand it.  I don’t deny this claim.  It may be that I don’t see "the point" to much of it.  I’m not sure.  Analysis, to me, is like a rigid building where everything must be just so — I contrast this with topology, where things seem somehow more fluid and less strict…

But enough whining.  These posts will go through the various topics in analysis, perhaps stopping along the way to offer solutions to various related problems.  To be perfectly honest with the reader, the reason I am doing this is because I need to take a Qualifying exam in Analysis in the Fall and I need to review a lot of this material.  Therefore, if you find an error — and there may be some! — do not hesitate to tell me.

## Measures: The Size of Things.

In Topology, there is not usually a notion of size, in that even very "large" looking things can be homeomorphic to very "small" looking things — the real line and the unit open interval, for example, are homeomorphic but the real line is "much bigger looking" than the unit interval.  If topology ruled the world then things might get messy: if some fabric costs $5 for one yard, then via some homeomorphism you could potentially get an infinite number of yards of fabric for$5.  This is not a great way to run a business.  But notice that even in this example there is some notion of measurement: a yard is an interval of a specific size which (relativity and such aside) which does not change from place to place; indeed, a yard of fabric is the same "length" as a yard of fish.