Lim Sup: Life Lessons.

August 22, 2011

I’ve always felt a little uneasy with Lim Sup and Lim Inf’s.  Today while reading Rudin’s Complex book, I finally "got" it.


To find the lim sup of your sequence: take every convergent subsequence’s limit and find the sup of those limits.


This is, of course, exactly what the definition is, but it wasn’t "visual" to me until I thought about it this way.  Is there something you learned early on that you weren’t quite clear with until much, much later?


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.

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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. 

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Jordan’s Lemma.

August 9, 2011

[This post is for those of you who are already comfy with doing some basic contour integrals in complex analysis.]


So you’re sitting around, evaluating contour integrals, and everything is fine.  Then something weird comes up.  You’re asked to evaluate an integral that looks like

\displaystyle \int_{-\infty}^{\infty} e^{aix}g(x)\, dx

for g(x) is continuous.  Eek.  Don’t panic though, because Camille Jordan’s gonna help you out.


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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.



I’ve been a bit busy moving and packing my things, but I’ve tried to keep my problem-doing up to practice for my analysis qual.  One problem that seems to come up quite a bit in the complex analysis part of the exam is something like the following:


Question.  Suppose that f is entire which satisfies |f(z)| \leq A|z|^{k} + B for every z\in {\mathbb C} and every A,B > 0.  Prove that f is a polynomial of degree at most k.


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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.

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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.


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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." 


"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."


"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!"

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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.]

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