# 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. Commented Dec 7, 2019 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.

I'll try to give a solution based on the relevant exercises in Tao's "An Introduction to Measure Theory".

Exercise $$1.3.24$$: Show that a function $$f: \mathbb{R^d} \to \mathbb{C}$$ is measurable if and only if it is the pointwise almost everywhere limit of continuous functions $$f_n: \mathbb{R^d} \to \mathbb{C}$$.

proof: If $$f: \mathbb{R^d} \to \mathbb{C}$$ is measurable. Let $$\epsilon > 0$$.By exercise $$1.3.25$$, $$\forall n \geq 1$$, there is a measurable set $$E_n \subset \mathbb{R^d}$$ of measure at most $$\frac{1}{2^{n+1}}$$ s.t the function $$f \cdot 1_{B(0,n)}$$ is locally bounded outside of $$E_n$$. In particular, this implies that $$\exists M_n < \infty$$ s.t $$|f(x) \cdot 1_B(0,n)(x)| \leq M_n$$ on $$B(0,n)\setminus E_n$$. Then since $$\int_\mathbb{R^d} |f \cdot 1_{B(0,n) \setminus E_n}| \leq \int_\mathbb{R^d} M_n \cdot 1_{B(0,n) \setminus E_n} < \infty$$, the function $$f \cdot 1_{B(0,n) \setminus E_n}: \mathbb{R^d} \to \mathbb{C}$$ is absolutely integrable. By Theorem $$1.3.20$$, there exists a continuous, compactly supported function $$f_n$$ s.t $$||f \cdot 1_{B(0,n) \setminus E_n} - f_n||_L{^{1}}(\mathbb{R^d}) \leq \frac{1}{n \cdot 2^{n+1}}.$$ WLOG, we assume that $$f_n$$ is supported on $$B(0,n) \setminus E_n$$. By Markov's inequality and sub-additivity of the lebesgue measure, we have that: $$m(\{x \in B(0,n): |f(x) - f_n(x)| \geq 1/n\}) = m(\{x \in \mathbb{R^d}: |f(x) \cdot 1_{B(0,n)}(x) - f_n(x)| \geq 1/n \}) = m(\{x \in \mathbb{R^d}: |f(x) \cdot 1_{B(0,n) \setminus E_n}(x) - f_n(x)| \geq 1/n \} \cup \{x \in \mathbb{R^d}: |f(x) \cdot 1_{E_n \cap B(0,n)}(x)| \geq 1/n \}) \leq \frac{n}{n \cdot 2^{n+1}} + \frac{1}{2^{n+1}} = 1/2^n.$$

Now, we have shown that for every $$n \geq 1$$, there exists a continuous function $$f_n: \mathbb{R^d} \to \mathbb{C}$$ for which the set $$\{x \in B(0,n): |f(x) - f_n(x)| \geq 1/n\}$$ has measure at most $$1/2^n$$. Let $$A$$ be the set on which $$f_n \not\to f$$ pointwise. Then $$A = \cup_{m=1}^{\infty}\cap_{n=1}^{\infty}E_{n,m}$$, where $$E_{n,m} = \{x \in \mathbb{R^d}: |f(x) - f_n(x)| \geq 1/m \}$$. Note that for any $$m \geq 1$$, $$\cap_{n=1}^{\infty}E_{n,m} \subset \{x \in B(0,n): |f(x) - f_n(x)| \geq 1/n \}$$ for sufficiently large n and all $$n' \geq n$$, and thus $$m(\cap_{n=1}^{\infty}E_{n,m}) \leq 1/2^n$$ can be made arbitrarily small by sending $$n \to \infty$$, and we have $$m(A) = 0$$ as desired.

Conversely, if $$f$$ is the pointwise almost everywhere limit of continuous functions $$f_n: \mathbb{R^d} \to \mathbb{C}$$. Then $$f$$ is measurable by (i) and (iv) of exercise $$1.3.8$$.