# Hopefully Interesting

## October 29, 2009

### A Hands-On Proof of Lusin’s Theorem

Filed under: analysis,math.CA — Pietro @ 3:40 am

Today I’d like to share a cute proof I came up with. Actually this is one of those posts that have been in the oven for the last year and a half, while my wife and I have been busy getting married, travelling across Europe, moving into a new place, defending my MSc in Logic and then moving to Los Angeles. But, since I’m retaking basic Real Analysis as a requirement of the PhD program at UCLA, it seems adequate to revisit this little idea and finally post it.

The proofs I’ve seen of Lusin’s theorem go through Egorov’s theorem; that’s how Stein-Shakarchi, Folland and Wikipedia do it. I don’t feel it is a particularly elegant proof, and it produces a truncated version of Lusin’s theorem, which holds only for sets of finite measure. A simple $\sigma$-finiteness argument takes care of this shortcoming, but one is left with the feeling that Egorov’s theorem is our hammer, and we’re trying to see Lusin’s as a nail. Hence the motivation for a different proof. I hope you will also appreciate how there are no real obstacles in the route below; everything that needs to be done can be done nearly without thinking.

First, let us state the theorem at a fairly low level of generality.

Theorem. (Lusin) Let $E\subseteq \mathbb{R}^d$ be measurable, and $f : E \to \mathbb{R}$ a measurable function. Then, for each $\varepsilon > 0$ there is a closed set $F_\varepsilon \subseteq E$ such that $E \backslash F_\varepsilon$ has measure at most $\varepsilon$ and the restriction of $f$ to $F_\varepsilon$ is continuous. (In the induced topology, of course.)

If one weren’t taught to be scared of arbitrary measurable functions as very complex objects, the first thing to attempt would be to remove bits of $E$ where $f$ was discontinuous until none were left. Surprisingly, this very simple-minded approach turns out to work.

What are witnesses to the discontinuity of $f$? One possibility is points $x\in E$ such that there is some sequence $(x_n)$ in $E$ converging to $x$ but for which $f(x_n)$ does not converge to $f(x)$. However, measure theory is fundamentally countable in nature: we are only able to control processes that are repeated a countable number of times. On the other hand, we will not get anywhere by removing a countable number of points.

So what is another witness? Thinking back to the topological definition of continuity, we come upon the following: $f$ is not continuous if there is some open set $O \subseteq^\circ \mathbb{R}$ such that $f^{-1}(O) \nsubseteq^\circ\mathbb{R}$. So we would be all right if we could fix up all the inverse images $f^{-1}(O)$, for open sets $O$, to be open. In fact, by basic topology, it’s enough to fix up all the $f^{-1}(B)$ for basic open sets $B$, and $\mathbb{R}$ admits a countable base for its standard topology.

Well, how do you fix these up? Suppose $f^{-1}(O)$ is not open in $E$, and we want to make it open by removing bits of $E$; when would that work? We need to find an $E'\subseteq E$ such that $E'\cap f^{-1}(O) \subseteq^\circ E'$ — that is, such that there is $\mathcal{O}\subseteq^\circ \mathbb{R}^d$ with $E'\cap f^{-1}(O) = E' \cap\mathcal{O}$.

That’s easier than it sounds. If we just take any open set $\mathcal{O}$ containing $f^{-1}(O)$, the problem  is that $E\cap\mathcal{O}$ might contain points besides those in $f^{-1}(O)$. We fix this by removing $\mathcal{O}\backslash f^{-1}(O)$ from $E$ to form $E'$. Then we will clearly have $f^{-1}(O) = E'\cap\mathcal{O}$ and so $f^{-1}(O) \subseteq^\circ E'$.

The good news is, since continuity is preserved by restrictions, we are free to run the above procedure for as many $O$ as we want, and each time we do it we won’t be screwing up our previous work. In other words, if $\{O_n\}$ is a countable base for the topology of $\mathbb{R}$, and $\mathcal{O}_n$ is an open set containing $f^{-1}(O_n)$, then removing $\mathcal{O}_1\backslash f^{-1}(O_1)$, $\mathcal{O}_2\backslash f^{-1}(O_2)$ etc from $E$ will make $f$ successively “more continuous”, and finally the restriction of $f$ to $E\backslash \{ \bigcup_{n\in\mathbb{N}} \mathcal{O}_n\backslash f^{-1}(O_n)\}$ will be fully continuous.

Now we just need to make the $\mathcal{O}_n\backslash f^{-1}(O_n)$ very small, say of measure less than $\varepsilon/2^n$; then the sum of the measures of the removed bits will be at most $\varepsilon$. This choice of $\mathcal{O}_n$ is guaranteed by the definition of measurability or a standard theorem, depending on your approach. You can take “good outer approximations by open sets” as the definition of measurability, or you can take Caratheodory’s “good splittings” definition, and get good approximations by open sets as an easy theorem. Either way, the proof is finished.

One final nitpick: we haven’t proved that, after removing all the junk, we’re left with a closed set. In fact, we usually won’t be. But we’re certainly left with a measurable set, so we can use the standard theorem on good inner approximations by closed sets and be done.

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