## 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.

## Sequences of Measurable Functions.

Given some sequence of measurable functions, $\{f_{n}\}$, which converge to some function $f$, we can actually conclude that $f$ is measurable.  For me, this theorem was not at all obvious, though we will revisit it when we consider simple functions (and at that point the intuition will lead us towards a better picture).

Theorem.  Given a sequence of (Lebesgue) measurable functions $\{f_{n}\}$ which converge to some function $f$ pointwise (that is, for every point $x$).  Then $f$ is measurable.

Before we do the proof of this, note that there are a few ways to approach this.  We can much more easily prove that a monotonic sequence converging pointwise gives a measurable function (essentially just use the definition of measurable here) and we can try to tease out general sequences from this, but Kolmogorov spits out the proof in essentially one line.

Proof.  Let’s give the one line, and then we’ll try to tease out the proof.

$\displaystyle \{x\,|\, f(x) < c\} = \bigcup_{k}\bigcup_{N}\bigcap_{n>N}\left\{x\, | \, f_{n}(x) < c - \frac{1}{k}\right\}$

That’s right.  Do we even need to go on?  Isn’t it obvious?  Didn’t you instantly think of this?  For those of you who didn’t, let’s go over it in a bit more detail.

The left side is just our usual set for checking measurability of $f$.  Nothing crazy.  The right side is where we get really nuts.  What is all this non-sense?  Let’s read it out: $x$ is in the right-hand-side if there exists some $k$ (this is the first union, over $k$) such that there exists some $N$ (this is the second union, over $N$) such that for every $n > N$ we have that $x$ is in the set $\left\{x\, | \, f_{n}(x) < c - \frac{1}{k}\right\}$.  Whew.

So the main part of the proof is actually showing that these sets are equal.  So let’s do that.  If $x$ belongs to the left-hand-side, then it is such that $f(x) < c$.  Then there should be some $k$ such that $f(x) < c - \frac{2}{k}$, just by the nature of real numbers (this 2 on top will make sense in a second!).  That’s no big deal.  But by the nature of convergence pointwise, we have that there exists an $N$ such that for all $n > N$ we will have $f_{n}(x)$ a distance at most $\frac{1}{k}$ away from $f(x)$, and hence $f_{n}(x) < c$ (that’s why we had the 2 above!) for every $n > N$.  This means $x$ belongs to the right-hand-side.

Okay.  Now.  If $x$ belongs to the right-hand-side, it means that there is some $k$ and some $N$ such that for every $n > N$ we have $f_{n}(x) < c - \frac{1}{k}$.  Thus we must have that $f(x) < c$ since the $f_{n}$ converge to $f$ pointwise.

Notice, though, that we know the sets on the right side are measurable, since each of the $f_{n}$ are measurable.  We also know that countable unions and intersections of measurable sets are measurable.  But this means that the entire right-hand-side (and therefore the left-hand-side) is measurable!  This shows that $f$ is measurable.  Nice.  $\Box$.

This theorem is pretty strong.  We can now build measurable functions out of sequences of other measurable functions.  But how should we begin to build measurable functions?

Well, how do people build anything?  We’ll start with the simplest things we can, and then put a bunch of them together to make something more complicated.  And the simplest kinds of functions are ones which are constant or “almost constant”; that is, ones which take on only finitely many values.

Definition.  We will call a Lebesgue measurable function $f$ a simple function if it only attains finitely many range values.

Notice the qualification that $f$ must be Lebesgue measurable.  For example, the characteristic function of a non-measurable set only takes on finitely many range values, but it is not simple since it is not measurable.  On the other hand, a step function or the characteristic function of the rationals is a simple function.

Here is an unsurprising corollary from the theorem above.

Unsurprising Corollary.  Given $\{f_{n}\}$ simple functions which converge uniformly to $f$, then $f$ is measurable (but not necessarily simple).

The addition of “uniform convergence” makes this assumption significantly stronger than the pointwise convergence which was needed above.  So this corollary should not surprise you.  If it does, then maybe you are easily surprised and you should get that checked out by a doctor.

On the other hand (and you may want to sit down for this!), every bounded measurable function can be approximated by a monotonically increasing sequence of measurable functions.  WHAT.  Yes.  Really.  (Note: The “bounded” part here is due to the fact that unbounded functions behave badly when it comes to uniform convergence.)

Theorem.  If $f$ is a bounded measurable function, then there exists a sequence of simple functions $\{f_{n}\}$ which converge uniformly to $f$.

I won’t lie to you, readers, this requires a trick.  And, in fact, this trick is so important that I will give it a name.  I call this the $\frac{{\bf m}}{{\bf n}}$ trick.  Let’s see how it works.

Proof.  Define $f_{n}(x) = \frac{m}{n}$ whenever $\frac{m}{n} \leq f(x) < \frac{m+1}{n}$.  We clearly have that $f_{n}(x) \rightarrow f(x)$ (if this isn’t clear, look at what value $f_{n}(x)$ is and in what range the $f(x)$ can be; $f(x)$ essentially gets “squished” between these two values, and $f_{n}(x)$ goes to this value as $n$ increases).  In addition, and here is the clever part, notice that for every $x$ it is the case that

$|f(x) - f_{n}(x)| < \frac{1}{n}$

which gives us convergence.  The fact that $f$ is bounded between, say, $[m,M]$, gives uniformity (why?).  And that’s the $\frac{m}{n}$ trick.  $\Box$.

Note that, generally, a measurable function will only be the pointwise limit of such functions — the proof is essentially the same as above.

So, we have the following theorem which actually gives us one extremely nice characterization of measurable functions:

Theorem.  A bounded function $f$ is measurable if and only if it is the limit of a uniformly convergent sequence of simple functions.  A (not necessarily bounded) function is measurable if and only if it is the pointwise limit of a sequence of simple functions.

But what if we don’t have simple functions?  If our sequence converges to some measurable function, is there anything we can say about it that was as nice as that $\frac{m}{n}$ trick?  Well, not quite.  I’ll save this for the next post, but suffice it to say that we can make such a sequence almost uniformly converge in the sense that it will converge uniformly except for little pieces.  The exact statement of this is an important theorem in measure theory called Egorov’s Theorem.  And you know it’s important when it’s named after someone!

### 4 Responses to “Real Analysis Primer, Part 7: Measurable Functions and Limits.”

1. Anonymous said

Work practice problems . . . .

2. Anonymous said

Not every measurable function is a uniform limit of simple functions. Bounded measurable functions are uniform limits of simple functions but unbounded functions are JUST point wise limits of simple functions.

3. Ilya said

In the above proof of measurability of a limit of a sequence of measurable functions, I’m not sure you need the 1/n. If you omit the 1/n. I believe strict inequality < c should give you the same solution.