# Lebesgue measurable function is a limit of continuous functions almost everywhere

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?

• A function $\mathbb{R}^n\rightarrow\overline{\mathbb{R}}$ is Lebesgue-measurable iff it is a.e. equal to a Borel-measurable function. – Thorgott Dec 7 '19 at 13:23

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|

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.