Lebesgue measurable function is a limit of continuous functions almost everywhere

Question

Show that for a Lebesgue measurable function $f\colon R^n\rightarrow R$, there is a sequence of continuous function $\{f_i\}$ that $f=\lim_{i\rightarrow\infty}f_i$ almost everywhere.

What about also prove the other side? i.e. if $f=\lim_{i\rightarrow\infty}f_i$ almost everywhere then $f$ is Lebesgue measurable.

A similar question can be found here. But it doesn't show how actually is this question proved. Is every Lebesgue measurable function on $\mathbb{R}$ the pointwise limit of continuous functions?

Answer 1

In fact the similar question you posted answers your question.

The only thing you need is that Luzin's Theorem is true for the whole space,not only for sets if finite measure.

Lets prove this.

We have that $\Bbb{R}^n=A_1 \cup \bigcup_{n=1}^{\infty}A_n$ where $A_0=\{x:|x|<1\}$ and $A_n=\{x:n \leq |x|<n+1\}$

Let $\epsilon>0$. Then exists a closed $F_n \subseteq A_n$ and $g_n$ continuous on $F_n$ such that $m(A_n \setminus F_n)<\frac{\epsilon}{2^{n+1}}$ and $g_n=f$ on $F_n$

Define $F=\bigcup_{n=0}^{\infty}F_n$ and $g=\sum_{n=0}^{\infty}g_n1_{F_n}$

$g$ is continuous on $F$(exercise)

$F$ is closed (exercise)

Since $F$ is closed we can extend $g$ to a continuous $G$ on the whole space by Tietze's theorem and $G=f$ on $F$ and $m(\{G \neq f\}) \leq m(\Bbb{R}\setminus F)<\epsilon$

Now $\forall n \in \Bbb{N}$ exists $G_n$ continuous on $\Bbb{R}^n$ such that $m(\{G_n \neq f\})< \frac{1}{2^n}$

By Borel-Cantelli we have that $m(\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}\{G_k \neq f\})=0$

Thus for almost every $x$ we have that exists $m \in \Bbb{N}$ such that $G_n(x)=f(x),\forall n \geq m$

So you have the conclusion.

For the other part,note that every continuous function is measurable.

Thus $\limsup_nf_n$ is measurable so $f=\limsup_n f_n$ is measurable.

Answer 2

Refer to Exercise $24$ and related results in Math 245A Note 2 for this problem.

Exercise $24$: Show that a function ${f: {\bf R}^d \rightarrow {\bf C}}$ is measurable if and only if it is the pointwise almost everywhere limit of continuous functions ${f_n: {\bf R}^d \rightarrow {\bf C}}$. (Hint: if ${f: {\bf R}^d \rightarrow {\bf C}}$ is measurable and ${n \geq 1}$, show that there exists a continuous function ${f_n: {\bf R}^d \rightarrow {\bf C}}$ for which the set ${\{ x \in B(0,n): |f(x)-f_n(x)| \geq 1/n \}}$ has measure at most ${\frac{1}{2^n}}$. You may find Exercise $25$ below to be useful for this.)

proof: Let ${f: {\bf R}^d \rightarrow {\bf C}}$ be measurable, and $\varepsilon > 0$. By Exercise 25, there is a measurable set ${E \subset {\bf R}^d}$ of measure at most $\varepsilon$ outside of which $f1_{B(0,n)}$ is locally bounded. In particular, $f1_{B(0,n) \backslash E}$ is bounded and thus absolutely integrable. By Theorem 15, there exists a continuous, compactly supported $f_n$ such that $\|f1_{B(0,n) \backslash E} - f_n\|_{L^1({\bf R}^d)} \leq \varepsilon$. By monotonicity we have \begin{align*} &\frac{1}{n} \cdot m({\{ x \in B(0,n): |f(x)-f_n(x)| \geq 1/n \}}) \\ &\leq \frac{1}{n} \cdot m(\{x \in B(0,n) \backslash E: |f(x) - f_n(x)| \geq 1/n\}) + \frac{1}{n} m(E) \\ &\leq \|f1_{B(0,n) \backslash E} - f_n\|_{L^1({\bf R}^d)} + \frac{1}{n} m(E) \\ &\leq \varepsilon(1 + \frac{1}{n}). \end{align*} By choosing $\varepsilon \leq \frac{1}{(n+1)2^n}$, we get a continuous function ${f_n: {\bf R}^d \rightarrow {\bf C}}$ for which the set $A_n := {\{ x \in B(0,n): |f(x)-f_n(x)| \geq 1/n \}}$ has measure at most $1 / 2^n$. Let $F := \{x \in {\bf R}^d: f_n(x) \not\rightarrow f(x)\}$. For every $x \in F$, there exists $N > 0$ such that $x \in A_n$ for all $n \geq N$. Hence $F \subset \bigcup_{N = 1}^\infty \bigcap_{n \geq N} A_n$, and we get $m(F) = 0$.

Conversely, if ${f: {\bf R}^d \rightarrow {\bf C}}$ is the pointwise almost everywhere limit of continuous functions ${f_n: {\bf R}^d \rightarrow {\bf C}}$, then $f$ is measurable by $(1)$ and $(4)$ of Exercise $8$.