# Measure Theory – When is Measurability Equivalent to Limit of Measurable Functions?

analysisborel-measuresmeasurable-functionsmeasure-theorysolution-verification

I have a suspicion that the following is true: $$\def\AAA {\mathcal{A}} \def\BBB {\mathcal{B}} \def\NN {\mathbb{N}}$$

Theorem: let $$(X,\AAA,\mu)$$ be a complete measure space, and $$(Y,\BBB)$$ a measurable space where $$\BBB$$ is the Borel $$\sigma$$-algebra of some metrizable topology of $$Y$$. Then, the following are equivalent:

1. $$f:X\to Y$$ is measurable.
2. $$f$$ is almost everywhere the pointwise limit of measurable functions $$f_n$$.

What I do know.

• The result holds if $$f_n\to f$$ everywhere. For a proof, see here.
• The result does not hold if the domain space is not complete (a counterexample can be constructed from this post).

My attempt.

I have been trying to modify this proof so as to deal with the case where $$f_n\to f$$ only almost everywhere. I first partition $$X$$ into
$$X_* := \{ x\in X : f_n(x)\to f\} \ \ \ \ \text{ and } \ \ \ \ X_0 := X\setminus X_*.$$
For any closed $$C\subseteq Y$$, completeness of $$X$$ allows us to claim
$$\Bigg(f^{-1}(C)\Bigg) \triangle \Bigg(\bigcap_{n\in\NN}\bigcup_{N\in\NN}\bigcap_{k\ge N}f_k^{-1}(C_{2^{-n}})\Bigg)$$
is a null set, call it $$C_0$$. As the set on the right, call it $$M$$, is measurable by hypothesis, we have
$$\big(f^{-1}(C)\setminus M\big)\cup\big(M\setminus f^{-1}(C)\big) = C_0$$
which hopefully, somehow, implies $$f^{-1}(C)$$ is measurable as well.

Is the 'Theorem' correct? If so, how could one prove it? If not, what is a counterexample?

Related.

• This post. I struggle to understand large part of the terminology, so it hasn't been of much help.

Yes, the theorem is correct. Let us prove it: $$\def\AAA {\mathcal{A}} \def\BBB {\mathcal{B}} \def\NN {\mathbb{N}}$$

Theorem: let $$(X,\AAA,\mu)$$ be a complete measure space, and $$(Y,\BBB)$$ a measurable space where $$\BBB$$ is the Borel $$\sigma$$-algebra of some metrizable topology of $$Y$$. Then, the following are equivalent:

1. $$f:X\to Y$$ is measurable.
2. $$f$$ is almost everywhere the pointwise limit of measurable functions $$f_n$$.

Proof:

(1 $$\Rightarrow$$ 2) Just take $$f_n=f$$ for all $$n$$.

(2 $$\Rightarrow$$ 1) Since $$f_n$$ are measurable functions, $$f_n$$ converges a.e. to $$f$$ and $$(X,\AAA,\mu)$$ is a complete measure space, there is a set $$E \in \AAA$$ such that $$\mu(E)=0$$ and $$f_n$$ converges to $$f$$ in $$X \setminus E$$.

Let $$g_n = \chi_{X\setminus E} f_n$$. It is clear that $$\chi_{X\setminus E}$$ is a measurable function and so $$g_n$$ are measurable functions.

It is easy to see that $$g_n$$ converges everywhere to $$\chi_{X\setminus E} f$$.

Since $$(Y,\BBB)$$ is a measurable space where $$\BBB$$ is the Borel $$\sigma$$-algebra of some metrizable topology of $$Y$$, we know that the pointwise (everywhere) limit of a sequence of measurable functions is a measurable function, so we have that $$\chi_{X\setminus E} f$$ is measurable.

Since $$f = \chi_{X\setminus E} f$$ a.e., $$\mu(E)=0$$ and $$(X,\AAA,\mu)$$ is a complete measure space, we have that $$f$$ is measurable.