The problem I am about to pose is a slightly weaker version of the exercise 18 (p. 42) of Stein's and Shakarchi's Real Analysis
Every measurable function is the limit a.e. of a sequence of continuous functions.
Problem statement: Let $\mu$ be a Borel regular outer measure on a topolopgical space $X$ and $f:X\to\mathbb{R}$ be a $\mu$-measurable function such that $|f(x)| < \infty$ outside a set of $\mu$-measure zero. Show that there exists a sequence of continuous functions which converges to $f$ outside a set of measure zero
My problem(s): By application of Lusin's theorem and Tietze's extension theorem, one will obtain the following: For every $n\in\mathbb{N}$, there exists a closed set $C_n\subset X$ such that $\mu(X\setminus C_n) < \frac{1}{n}$ and there exists a sequence of continuous functions $(F_{n,k})_{k=1}^\infty, F_{n,k}:X\to\mathbb{R}$ such that $\lim_{k\to\infty}\sup_{x\in C_n}|f(x) – F_{n,k}(x)| = 0$, i.e. $F_{n,k}$s converge uniformly on $C_n$ to $f$.
All this is equivalent to saying that $\mu$ almost everywhere there exists a sequence of continuous functions converging to $f$. But the problem is asking more, that there exists a sequence of continuous functions converging a.e. to $f$.
With this in mind, I took the function $f_n$ to be such $F_{n,k_n}$ that $\sup_{x\in C_n}|f(x) – F_{n,k_n}(x)| < \frac{1}{n}$ and defined $C := \bigcup_{n=1}^\infty C_n$. Then $\mu(X\setminus C) < \frac{1}{m},\forall m\in\mathbb{N}$ and so $\mu(X\setminus C) = 0$. But while $F_{n,k_n}$ behaves well on $C_n$, I don't see any reason why $|f(x) – F_{n,k_n}(x)|,x\in C\setminus C_n$ should be particularly small for any $n\in\mathbb{N}$. Therefore if I am not mistaken, this sequence will not do the job.
Is there any hope of salvaging this proof? How should I choose the sequence of continuous functions?
Best Answer
Your sequence works if we additionally assume that $C_n \subset C_{n+1}$. Indeed, if $x \in C=\bigcup_{n=1}^{\infty} C_n$, we have that $x \in C_n$ for all $n \geq m$ for some $m \in \mathbb{N}$. Hence, for $n \geq m$ $$|f(x)-F_{n,k_n}(x)| \leq \frac{1}{n}.$$
(In fact, since Lusin's theorem gives that $f$ is continuous on $C_n$ we could simply take $F_n$ a continuous extension of $f|_{C_n}$, for which $\sup_{x \in C_n}|f(x)-F_n(x)|=0$ holds.)
The reason why we may assume that $\{C_n\}_n$ is increasing, is because $f$ is continuous on each $C_n$ (Lusin), and hence, also continuous on finite unions. Therefore, we may redefine $\widetilde{C}_n = \bigcup_{k=1}^n C_k$.