# Every measurable function is the limit a.e of a sequence of continuous functions

Problem. Prove that every measurable function is the limit a.e of a sequence of continuous functions

• "a.e" mean almost everywhere

• Here, $$m$$ is the Lebesgue measure

Idea. Let $$\psi = \sum_{1}^{n}a_{j}\chi_{R_{j}}$$ be a step function where $$R_{j}$$ are closed rectangles. Each $$R_{j} \subset \mathbb{R}^{d}$$ is of the form $$R_{j} = \prod_{1}^{d}[a_{i},b_{i}]$$. For $$\chi_{[a_{i},b_{i}]}$$ consider $$g_{[a_{i},b_{i}]}(x) = \begin{cases} 1,& x \in [a_{i},b_{i}]\\ 0,& x \geq b_{i} + \epsilon\;\mathrm{ou}\;x\leq a_{i} - \epsilon\\ \frac{1}{\epsilon}(x - a_{i} + \epsilon), & x \in (a_{i}-\epsilon,a_{i})\\ 1 - \frac{1}{\epsilon}(x - b_{i}),& x \in (b_{i},b_{i} + \epsilon) \end{cases}$$ that is continuous and igual to $$\chi_{[a_{i},b_{i}]}$$ a.e. Thus, for $$R_{j}$$, $$\chi_{R_{j}} = g_{R_{j}}$$ a.e. Since $$\psi$$ is finite linear combination of $$n$$ characteristics functions defined in rectangles, then $$\psi = \sum_{1}^{n}a_{j}g_{R_{j}}$$ a.e.

Now, let $$f$$ be a measurable function. Then there is a sequence of step functions tht converges pointwise to $$f$$ a.e. Let $$g_{k}$$ be a continuous function with $$\psi_{k} = g_{k}$$ a.e. Let $$Y$$ be a null set such that the sequence of step function doesn't converges to $$f$$ and $$X_{k}$$ a set of measure at most $$2^{-k}$$ such that $$\psi_{k} \neq g_{k}$$. Thus, if $$x \in \bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}(X_{k}^{c}\setminus Y)$$ then there is $$n$$ such that for $$k>n$$ we have $$\psi_{k}(x) = g_{k}(x)$$. Therefore, for this $$x$$, define $$f(x) = \lim_{k \to \infty}\psi_{k}(x) = \lim_{k \to \infty}g_{k}(x).$$ Now, define $$Z = \left(\bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}(X_{k}^{c}\setminus Y)\right)^{c} = Y \cup \left(\bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}X_{k}\right)$$ and so, $$\begin{eqnarray*} m(Z) & \leq & m(Y) + m\left(\bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}X_{k}\right)\\ & \leq & m\left(\bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}X_{k}\right)\\ & \leq & m\left(\bigcup_{n=k}^{\infty}X_{k}\right)\\ & \leq & \sum_{k=n}^{\infty}m(X_{k})\\ & \leq & \sum_{k=n}^{\infty}2^{-k} = 2^{1-n} \end{eqnarray*}$$ and when $$n \to \infty$$, $$m(Z) = 0$$. For $$x \not\in Z$$, $$g_{n} \to f$$ pointwise.

This is the idea that I know to prove it, but I'm trying to get a different proof, a little more intuitive maybe. Somebody knows?

• This is more or less "Lusin's theorem". Sep 27, 2018 at 23:29

Well, let $$f$$ be a measurable function, without loss of generality you can consider $$f\geq 0$$, now you can approximate $$f$$ by simple functions. Now, between them, in the possible discontinuities, cut an $$\epsilon=2^{-k-1}$$ (where k is the number of the addens in the simple aproximation) and glue by a line making them continuous. Now, you have to play with the epsilon and you are ready.

Actually we can find a sequence of $$C^{\infty}$$ functions with compact support converging a.e. to $$f$$. Since $$(\tan ^{-1} f) I_{(-n,n)}$$ is an inegrable function there exists a $$C^{\infty}$$ function $$g_n$$ with compact support such that $$\int |(\tan ^{-1} f) I_{(-n,n)}-g_n|\, dx <\frac 1 n$$. [ See Theorem 3.14 in Rudin's RCA].This implies that a subseqeunce of $$\{(\tan ^{-1} f) I_{(-n,n)}-g_n\}$$ converges almost everywhere. From this the result follows immediately. [Just consider $$\tan g_n$$ along the subsequence].

• Nice! It's a very different approach! Sep 29, 2018 at 1:00
• Great proof. I think Rudin only proves that $C_c$ is dense in $L^p; p<\infty$, though. The $C_c^{\infty}$ case requires more work. Sep 29, 2018 at 1:40
• I am using Rudin's theorem with p=1. Sep 29, 2018 at 4:39

As in the other answer, the $$f_n$$ can be constructed. Here is a different approach.

Since $$\mathbb R^d$$ is $$\sigma-$$ finite and since there is a sequence of simple functions (not necessarily continuous) converging to $$f$$, it suffices to prove this for $$f=\chi_E$$, the indicator function of a measurable set $$E$$ with $$\ \lambda (E)<\infty.$$

Using regularity of the Lebesgue measure, we find compact sets $$K_n\subset E$$ and open sets $$U_n\supseteq E$$ such that $$\lambda(U_n-E)<1/n$$ and $$\lambda(E-K_n)<1/n.$$

Now, $$U^c_n$$ and $$K_n$$ are closed and disjoint, so we may apply Urysohn's Lemma, to produce a sequence $$(f_n)$$ of continuous functions that satisfy $$0\le f_n\le 1,\ f_n(K_n)=1$$ and $$f_n(U_n)=0.$$ Then, $$|\chi-f_n|=0$$ except on $$U_n- E$$ and $$E-K_n$$ each of which is aset of maeasure less than $$1/n,$$ which implies that $$f_n\to \chi_E$$ except on $$\bigcap_n (U_n-E)\cup(\bigcup_n E-K_n)$$, which has Lebesgue measure zero.

• Interesting! It's the first time I see Urysohn's Lemma applied to a question. Sep 29, 2018 at 0:59
• @LucasCorrêa: maybe Urysohn is overkill, because you can use $f_n(x)=\frac{d(x,U_n^c)}{d(x,K_n)+d(x,U_n^c)}$. Sep 29, 2018 at 1:17
• I liked. I studied the lemma and had not used it anywhere until now. Sep 29, 2018 at 1:24