When is measurability of a function $f$ equivalent to $f$ being almost everywhere the limit of measurable functions?

Question

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 $f_n$.

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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).

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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.

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

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Related.

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

Answer 1

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.