Notes on Luzin and Egoroffs' Theorems

These are some notes on Luzin and Egoroffs’ theorems. These are theorems are famous for being formalizations of two of Littlewood’s principles:

• Every measurable function is nearly continuous.
• Every pointwise convergent sequence of measurable functions is nearly uniformly convergent.

For reference, I’m using Royden and Fitzpatrick (R&F). I present proofs of both theorems that are not in R&F, and discuss them. First, here is a theorem that is essential for R&Fs’ proof that is also essential to mine.

Theorem (R&F 2.11). Each of the following four assertions is equivaent to the measurability of the real set $E$.

1. For each $\epsilon > 0$, there is an open set $\mathcal O$ containing $E$ for which $m^*(\mathcal O \setminus E) < \epsilon$.
2. There is a $G_\delta$ set $G$ containing $E$ for which $m^*(G\setminus E) = 0$.
3. For each $\epsilon > 0$, there is a closed set $F$ contained in $E$ for which $m^*(E\setminus F) < \epsilon$.
4. There is an $F_\sigma$ set $F$ contained in $E$ for which $m^*(E\setminus F) = 0$.

Theorem (Luzin). Let $f$ be a real-valued measurable function on $E$. For each $\epsilon > 0$, there is a continuous function $g$ on $\mathbb{R}$ and a closed set $F$ contained in $E$ for which \begin{equation} f = g \text{ on } F \text{ and } m(E\setminus F) < \epsilon. \end{equation}

Royden’s proof is broken into the following steps:

1. Prove the statement when $f$ is a simple function.
   1. Use item 3. of R&F 2.11 to produce a collection of closed sets where $f$ only takes one value.
   2. There is a continuous way to connect these closed sets, i.e. with linear connections.
2. Prove the statement for all functions.
   1. Consider a sequence of simple functions ${f_n}$ that converge pointwise to $f$. Applying the statement for simple functions, we can get a sequence of continuous functions ${g_n}$ and closed sets ${F_n}$.
   2. If ${F_n}$ are defined with the right tolerance $\epsilon_n$, we can define $F’ = \cap_{n=0}^\infty F_n$ and $m(E\setminus F’) < \epsilon/2$.
   3. We can use Egoroff’s theorem to identify $F \subset F’$ where ${g_n}$ converges uniformly, and $m(E\setminus F) \leq \epsilon$.
   4. On $F$, $g_n=f_n$. The uniform limit of continuous functions is continuous. As ${f_n} \rightarrow f$, $f$ will be continuous on this set.
   5. We can define $g$ to extended $f$ to the real line and be continuous, i.e. with linear connections.

In total, the proof stretches about a page, not including the two items which are left as exercises for the reader. Intuitively, it leaves something to be desired, at least from me. It hinges on approximating $f$ by a sequence of simple functions. Then, it produces that $f$ is continuous by leveraging this lesser-known fact that the uniform limit of continuous functions is continuous. In doing so, we’re using flat simple functions to a approximate a curvy $f$, and saying that the limit is in fact curvy. The key, basically, is that as we take increasingly more-resolved step-type approximations of $f$, sort of `buffer’ intervals are taken around each jump, but they are created in a way that their total measure is finite. So, if we were to follow this proof in a constructive way, the set $F$ would have a complicated structure, actually rather like a generalized Cantor set (see R&F, problem 2.39), at least in the sense that it is dense with positive measure. That also sort of leaves something on the table. Even for letting $f$ be continuous, $F$ as constructed by this proof will have this complicated structure, where it in fact could be all $E$.

Luzin’s theorem is a realization of Littlewood’s intuitive principle that measurable functions are nearly continuous. And intuitive things generally have quick proofs that work naturally with the developed machinery. In particular, we could examine the equivalent definition of a measurable function.

Proposition (R&F 3.2). Let the function $f$ be defined on a measurable set $E$. Then $f$ is measurable if and only if for each open set $\mathcal O$, the inverse image of $\mathcal O$ under $f$, $f^{-1}(\mathcal O) = \{x \in E \vert f(x) \in \mathcal O\}$, is measurable.

The definition only differs from that of a continuous function in that the preimage is measurable rather than open. But measurable sets are nearly open (R&F 2.11, 1), so we should expect that measurable functions are nearly continuous. This would be the line of attack I would chose for a proof of Luzin’s theorem, and is what I give below. A cursory search shows that it is not original; this arxiv paper reports essentially the same idea, and they also do not claim to be the originators. That arxiv paper gives the proof in the Borel setting, which gives me confidence that generalizability is not the reason the textbook authors do not use this proof. Rather, I guess they do it for the purpose of instruction. In one way, the ability to create super complicated closed sets is a superpower of measure theory, and familiarity with proof techniques which utilize that will be useful to the student. In another sense it complicates intuition behind the theorem.

Proof. Let $ \left\{a_i, b_i\right\}_{i=1}^\infty$ be an enumeration of all intervals where $a_i$ and $b_i$ are rational. From (R&F 3.2), $f^{-1}((a_i,b_i))$ is measurable. By (R&F 2.11, 1.), there exists an open set $O_i \supset f^{-1}((a_i,b_i))$, and \begin{equation} m(O_i\setminus f^{-1}((a_i,b_i))) < \epsilon / 2^i. \end{equation} So, we can define $F = E\setminus \bigcup_{i=1}^\infty (O_i \setminus f^{-1}((a_i,b_i)))$. Then, using subadditivity, \begin{equation} m(E\setminus F) = m\left(\bigcup_{i=1}^\infty O_i \setminus f^{-1}((a_i,b_i))\right) \leq \sum_{i=1}^\infty m(O_i \setminus f^{-1}((a_i,b_i))) < \sum_{i=1}^\infty \epsilon / 2 ^i = \epsilon. \end{equation} Now we go on to show that $f\vert_F$ is continuous. Let $U$ be an open set; it is a countable union of rational intervals: \begin{equation} U = \bigcup_{n=1}^\infty (a_{i_n}, b_{i_n}). \end{equation} By construction, $f^{-1}((a_{i_n}, b_{i_n}))\cap F = O_{i_n}\cap F$ and \begin{equation} f\vert_F^{-1}(U) = \bigcup_{n=1}^\infty (O_{i_n}\cap F) = \left(\bigcup_{n=1}^\infty O_{i_n}\right)\cap F. \end{equation} An open set intersected with the space, $f\vert_F$ matches the definition of a continuous function.

What’s left is to show that a continuous function $g$ on $E$ can be constructed which is equal to $f$ on $F$. It’s not too hard and I’ll describe it in words. $F$ is closed, so $E \setminus F$ is open, and the countable intersection of open intervals. Those intervals may have $f$ defined on either end – in which case you can extend with linear interpolation, or with $f$ defined on one end and the interval goes to $\pm \infty$. Then you can extend with the value of $f$.$\Box$

Unlike the other proof, the $F$ constructed via this proof isn’t necessarily dense in $E$. For a continuous $f$, we could validly chose $O_i = f^{-1}((a_i,b_i))$, and then $F=E$. So, we come about the reasonable result that the space need not be modified where $f$ is already properly behaved. But, when $O_i$ truely surrounds $f^{-1}((a_i,b_i))$, you could imagine that the set $F$ could still become very complicated.

R&F use Egoroff’s theorem to prove Luzin’s. That precludes them from using Luzin’s to prove Egoroff’s, and we can ask if Egoroff’s theorem be produced more cleanly from Luzin’s. Below is one proof of Egoroff’s theorem after Luzin’s has been determined.

Theorem (Egoroff). Assume $E$ has finite measure. Let ${f_n}$ be a sequence of measurable functions on $E$ that converges pointwise on $E$ to the real-valued function $f$. Then for each $\epsilon > 0$, there is a closed set $F$ contained in $E$ for which \begin{equation} {f_n} \rightarrow f \text{ uniformly on } F \text{ and } m(E\setminus F) < \epsilon. \end{equation} Proof. Fix $\epsilon > 0$. We’ll apply Luzin’s theorem on the measurable functions $\sup_{n \geq N}|f_n(x) - f(x)|$, in doing so getting continuous $g_N$ and closed $F_N$ so that \begin{equation} g_N(x) = \sup_{n\geq N}|f_n(x) - f(x)| \text{ on } F_N \text{ and } m(E\setminus F_N) < \frac{\epsilon}{2^N}. \end{equation} Define $F = \cap_{N=1}^\infty F_N$, which is closed, and by subadditivity $m(E\setminus F) < \epsilon$. Fix $\epsilon’ > 0$ (used to show uniform convergence). Look at the open cover ${E_N}$ of $F$, where \begin{equation} E_N = \{x\in E \,\vert\,\, g_N(x) < \epsilon’ \}. \end{equation} $E_N$ are open because $g_N$ is continuous, and $E_N$ cover $F$ because $g_N$ converges pointwise to zero on $F$. Because $m(E) < \infty$, so too $m(F) < \infty$, and the Heine-Borel theorem can be used to extract a finite subcover, $\{E_{N_i}\}$. Define $\bar N = \max_{i} {N_i}$. Because $g_N$ is decreasing (with $N$), $g_{\bar N} < \epsilon’$ everywhere. And by the definition of $g_N$, $|f_n(x) - f(x)| < \epsilon’$ for all $n \geq \bar N$. $\Box$

The second half of this proof is in effect the proof of Dini’s Theorem. The proof also constructs $f$ to be continuous on $F$, which is not stated in the theorem, but would be immediate from an application of Egoroff’s, then Luzin’s theorems anyways.

Egoroff’s theorem is closely related to the fact that pointwise convergence implies convergence in measure (PC $\Rightarrow$ CIM). The proof of PC $\Rightarrow$ CIM is direct from Egoroff’s theorem. At the same time, R&F’s proof of Egoroff’s first requires a proof of the following Lemma, which is practically just PC $\Rightarrow$ CIM.

Lemma (R&F 3.10) Under the assumptions of Egoroff’s Theorem, for each $\eta > 0$ and $\delta > 0$, there is a measurable subset $A$ of $E$ and an index $N$ for which \begin{equation} \left|f_n - f\right| < \eta \text{ on } A \text{ for all } n \geq N \text{ and } m(E\setminus A) < \delta. \end{equation}

Perhaps the above should be thought of as the simplest version of Egoroff’s idea. After this, Egoroff’s theorem is just that we can take $\eta \rightarrow 0$, while shrinking $\delta$ fast enough so that the intersection of all $A$’s is still nearly $E$.

Here’s a very simple statement that gets to the same idea. If $\{f_n\} \rightarrow f$, measurable functions, then for every $\eta$, $N$, the set \begin{equation} \{x \,\vert\,\, \max\{n \vert\,\,|f_n(x) - f(x)| \geq \eta\} = N \} \text{ is measurable.} \end{equation}

This is the set of points that dip within $\eta$ of $f(x)$ at $N$ and stay within $\eta$ ever after. Why is it measurable? Because if the above set is $D(\eta, N)$ \begin{equation} D(\eta,N) = \{x \,\vert\,\, |f_N(x) - f(x)| \geq \eta \}\cap \bigcap_{n > N} \{x \,\vert\,\, |f_n(x) - f(x)| < \eta \}. \end{equation}

Now, under the assumptions of Egoroff’s theorem, for any $\eta$, $\{D(\eta,N)\}_{N=1}^\infty$ partition $E$. Chose $N(\eta)$ large enough so that

\begin{equation} m\left(E\setminus\bigcup_{l=1}^{N(\eta)} D(\eta, l)\right) = m(E) - \sum_{l=1}^{N(\eta)} m(D(\eta, l)) < \frac{\epsilon}{2^{1/\eta+1}}. \end{equation}

Let the sequence $\{\eta_r\} = \{1/r\}$. Then, set $F’= \bigcap_{\eta_r}\bigcup_{l=1}^{N(\eta_r)}D(\eta_r,l)$. Manifestly, $m(E\setminus F’) < \epsilon/2$, and for any $\epsilon’ > 0$, there is an $r$ and $1/r < \epsilon’$. $\bigcup_{l=1}^{N(\eta_r)}D(\eta_r,l) \subset F’$, so for $n > N(1/r)$, $\vert f_n - f\vert < 1/r < \epsilon$ on $F’$. At the cost of an addition $\epsilon/2$, $F’$ can be closed using (R&F 2.11, 3.) to create $F$.

At the same time, because $\{D(\eta, N)\}_{N=1}^\infty$ are disjoint, for any $\eta > 0$, $\epsilon > 0$, we can find an $N$ so that for $n > N$,

\begin{equation} m\left(\{x\,\vert\,\, \vert f_n(x) - f(x)\vert > \eta \}\right) \leq m\left(E\setminus\bigcup_{l=1}^N D(\eta,l) \right) = m(E) - \sum_{l=1}^N m(D(\eta, l)) \leq \epsilon/2 < \epsilon. \end{equation}

so PC $\Rightarrow$ CIM.