UPDATE: I added an answer based off the hints provided by Robert Israel. It may, however, need some adjustment.

I'm trying to solve #18 (pg. 42) from Stein and Sharkarchi's analysis text.

(Before reading, I added the if part (<=) to the question since it didn't seem like much extra work to prove the statements were equivalent).

Problem Statement -- A function is measurable if and only if it is the a.e. limit of continuous functions.

My proof: Assume $\{f_{n}\}\to f\;a.e.\;x$ is a sequence of continuos functions. Then by property 2, each $f_{n}$ is measurable, and by property 4, their $a.e.$ limit is measurable. Thus, $f$ is measurable. Now suppose $f$ is measurable, and further assume $f$ is defined on a set $A\subset\mathbb{R}^{d}$ of finite measure and is infinite only on a set of measure zero. Let $B\subset A$ be the set where $f$ is finite so that $m(A-B)\leq\epsilon\;\forall\epsilon>0$. Then Lusin's Theorem guarantees the existence of a compact set $C\subset B$ where $m(B-C)\leq\epsilon\;\forall\epsilon>0$ and $f_{C}$ uniformly continuous. An application of the Tietze Extension Theorem allows us to construct a new function $F$ such that $F = f_{C}$ on $C$ and is continuous on all of $B$. Define a sequence of functions on $B$ by $f_{n} = F\;\forall\;n=1,2\ldots$ Then $\{f_{n}\}$ is (trivially) a sequence of continuous functions which uniformly converges to $F$ on $B$ (to obtain a non-trivial sequence of continuous functions, one could apply the generalized Stone-Weirstrass Approximation Theorem to $F$). As consequence, $f_{n}\to f\;a.e.\; x\in B$, since $m(B-C)\leq\epsilon$. But the union of two sets of measure zero is again a set of measure zero, so in fact $f_{n}\to\; f\; a.e.\; x\in A$

I'm concerned that my proof strategy is off the mark, particularly since I construct a rather trivial sequence of continuous functions. Secondly, how can I strengthen the theorem by getting rid of the hypotheses of $A$ having finite measure and $f$ being finite? (I managed to make $f$ finite a.e., but I don't think that's strong enough?).


  • $\begingroup$ What properties are you referring to? $\endgroup$ – AD. May 9 '12 at 16:02
  • $\begingroup$ They are taken from Stein and Shakarchi's text. Property 2 proves every continuous function is measurable, and property 4 proves every limit of a sequence of measurable functions is measurable (including lim sup/inf, etc.), and by extension these properties being weakened to a.e. $\endgroup$ – Sargera May 9 '12 at 16:07
  • $\begingroup$ Hint: if $m(A) = \infty$, write $A = \bigcup_{n} (A \cap B_n)$ where $B_n$ is a ball of radius $n$. If $|f| = \infty$ on a set of positive measure, consider "cutoff" functions that take the values $\pm n$ where $f = \pm \infty$. $\endgroup$ – Robert Israel May 9 '12 at 16:26
  • $\begingroup$ Robert, Are you suggesting that I apply Lusin's Theorem to the "cutoff" functions $f_{N}$ on the set $A_{N}=\bigcup_{n=1}^{N}\left(A\cap B_{n}\right)\;\forall\;n=1,2,\ldots$? Then since I can construct the continuous sequence of functions converging to $f_{N}$ on $A_{N}$ for each $N$, it must be true in the limit as $N\to\infty$, and therefore true for $f$ on $A$? \\ Also, is the sequence of continuous functions I exhibited by using the Tietze Extension Theorem legitimate? Or do you suspect the authors had something less trivial in mind? $\endgroup$ – Sargera May 9 '12 at 17:31
  • 1
    $\begingroup$ @JTian I suggest that you write the propositions in the text. Most people don't have your book and it is not standard names either. $\endgroup$ – AD. May 9 '12 at 20:17

Here is a tentative solution. Thank you Robert.

Problem Statement: A function is measurable if and only if it is the a.e. limit of continuous functions.

Proof: The reverse direction is trivial (see original post).

To prove the converse, we begin with proving a sequence of weaker statements which lead to the conclusion (since we want to apply certain theorems which require stronger hypotheses on $f$, and also because it significantly simplifies the notational mess that would occur if we had to deal with all exceptional cases at once). First, suppose $f$ is measurable, $f$ is defined on a set $A\subset\mathbb{R}^{d}$ of finite measure, and $f$ is finite on $A$. Then Lusin's Theorem guarantees us the existence of a compact $B\subset A$ where $m(A-B)\leq\epsilon$ for all $\epsilon>0$ and $f_{B}$ is uniformly continuous ($f_{B}$ is the "restriction" of $f$ to $B$; it is not defined on $A$!). An application of the Tietze Extension Theorem allows us to construct a new function $F$ such that $F$ is continuous on all of $A$ and $F=f_{B}$ on $B\subset A$ ($F$ is the continuous extension of $f_{B}$ from the subset $B$ to $A$). There are now many ways to construct a sequence of continuous functions. One trivial example is $\{f_{n}\}_{n=1}^{\infty}$ where $f_{n}=F\;\forall\;n=1,2,3\ldots$. A less trivial one comes from an application of the Stone-Weierstrass Approximation Theorem which guarantees the existence of a sequence of continuous polynomials which converge uniformly to $F$ on any compact set (take the closure of $A$ if necessary). In any case, we have a sequence of continuous functions $\{f_{n}\}_{n=1}^{\infty}$ where $f_{n}\to F$ on $A$. And since $F=f_{B}$ on $B$ is equivalent to $F=f$ on $B$, we have that $f_{n}\to f\;a.e.\;x\in A$ since $m(A-B)\leq\epsilon$.

Now weaken the hypotheses slightly by allowing $f$ to be infinite on a set of measure zero. Then we can let $B\subset A$ be the set on which $f$ is finite, where $m(A-B)=0$. Then from the above proof, we see there is a sequence of continuous functions $\{f_{n}\}_{n=1}^{\infty}$ which converges to $f\;a.e.$ on $B$. Let $C\subset B$ be the subset on which the convergence occurs. Then $f_{n}\to f$ on $(A-D)$ where $D=(B-C)\cup(A-B)$ is the union of two sets of measure zero. Hence, $f_{n}\to f\;a.e.\;x\in A$ since $D$ being the union of two sets of measure zero is again a set of measure zero.

Weakening the hypotheses further, we now allow the possibility of $E$ having non-finite measure, in particular, $E$ can be unbounded. Then $A=\bigcup_{n=1}^{\infty}A_{n}$ where $A_{n}=A\cap B_{n}$ and $B_{n}$ is a ball of radius $n$ about the origin. Since $\{B_{n}\}_{n=1}^{\infty}\nearrow\mathbb{R}^{d}$, for any subset of $A$ you desire $a.e.$ convergence, there exists an $N$ such that an application of the above proofs yield the desired result on $A_{N}$, and letting $N\to\infty$ gives the required conclusion on all of $A$.

Finally, we weaken the hypotheses to the original statement. In particular, we now allow $f$ to be infinite on a set of positive measure. To prove the same result holds, consider "cutoff" functions $g_{n} = f$ on the set where $f$ is finite, and $g_{n}=n$ on the set where $f$ is infinite. Then each $g_{n}$ is measurable, and so the above proofs apply to $g_{n}$ for all $n=1,2,\ldots$ Evidently $g_{n}\to f$ as $n\to\infty$, and so the sequences of continuous functions converging $a.e.$ to each $g_{n}$, also converge to $f$ at points where $f$ is finite, and as $n\to\infty$, to arbitrarily large values at points where $f$ is infinite. We therefore conclude the theorem holds when $f$ is infinite on sets of positive measure, thus completing the proof. (Note, in the final two cases, the index $n$ has nothing to do with the index on sequences of functions; it in fact indexes a collection of sequences of functions). QED

NOTE: I'm a little unsure about the final two cases; I was trying to apply Robert's hints, but I think I'm not being technical enough to offer a rigorous proof. Modifications are welcomed.


I'm not sure if this solution works, but if it does, it seems much simpler than the other one given here:) Maybe we can attack this problem by considering the following two cases.

Let $f$ be a measurable function defined over a measurable set $E$.

i)$f:E\to\mathbb{R}$ measurable.
For each $n\in\mathbb{N}^+$, define $B_n$ to be the closed ball centered at the origin with radius $n$. $E\cap B_n$ is measurable and of finite measure. By Lusin's theorem, there exists $F_n$ closed with $F_n\subset E\cap B_n$, $m(E\cap B_n-F_n)<2^{-n}$ and $f$ continuous when restricted to $F_n$. By Tietze extension theorem, there exists continuous $f_n:E\to\mathbb{R}$ with $f_n = f$ when restricted to $F_n$. If sequence $f_n$ does not converge to $f$ at $x\in E$, then $x\in E-F_n$ for infinitely many $n$. $E-F_n \subset (E\cap B_n-F_n)\cup B_n^c$. But $x\in B_n^c$ for only finitely many $n$. Thus, $x\in E\cap B_n-F_n$ for infinitely many $n$. By Borel-Cantelli theorem, since $\sum_{n=1}^{\infty}m(E\cap B_n-F_n)<\infty$, $m(\limsup_{n\to\infty}E\cap B_n-F_n) = 0$. Thus, $m(x\in E:f_n\text{ does not converges to }f\text{ at }x) = 0$ and $f_n\to f$ almost everywhere.

Indeed, when applying Tietze extension theorem above, we can take $f_n$ such that $\sup\{|f_n(x)|:x\in E\} = \sup\{|f(x)|:x\in F_n\}\leq\sup\{|f(x)|:x\in E\}$ for all $n$.

Denote by $\overline{\mathbb{R}}$ the extended real line with the standard topology.
ii)$f:E\to\overline{\mathbb{R}}$ measurable.
Denote a homeomorphism from $[-1,1]$ to $\overline{\mathbb{R}}$ by $h$(for example, $h(x) = \tan(x\pi/2)$). Easy to see $(h^{-1}\circ f):E\to[-1,1]$ is measurable. From the first case, there exists sequence $g_n:E\to\mathbb{R}$ continuous with $g_n\to (h^{-1}\circ f)$ a.e. and $\sup\{|g_n(x)|:x\in E\}\leq\sup\{|(h^{-1}\circ f)(x)|:x\in E\}\leq 1$. We conclude that $h\circ g_n$ is a sequence of continuous functions from $E$ to $\overline{\mathbb{R}}$ and $(h\circ g_n)\to f$ a.e..


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