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This is the Proposition 2.11 of the book Real Analysis from Folland.

The following implications are valid if and only if the measure is complete:

(a) If $f$ is measurable and $f = g$ $\mu$-a.e., then $g$ is measurable.

(b) If $f_n$ is measurable for $n\in \mathbb{N}$ and $f_n\rightarrow f$ $\mu$-a.e., then $f$ is measurable.

The complete solution to this question can be found in this answer. But note that it was considered that $f,g:(X,\mathcal{M})\to(\mathbb{R},\mathcal{B}_{\mathbb{R}})$, other solutions like [1] or others that I found on the internet also assume that the image is $\mathbb{R}$ or $\overline{\mathbb{R}}$.

I'm thinking about the general case, where, $f:(X,\mathcal{M})\to(Y,\mathcal{N})$. In this answer, the proposition $\mu$ is complete $\Rightarrow$ (a) is demonstrated in the general case.

I tried to show the opposite direction but I couldn't, in the general case is this proposition true? How to demonstrate this?

Mrcrg
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    If you read the material just before the proposition 2.11 in Folland's, you will see that this proposition is about functions taking values in $\mathbb{R}$ (or $\overline{\mathbb{R}}$ or $\mathbb{C}$, the three versions of proof are essentially the same). That is what is meant in Folland's. On the other hand, if you consider functions taking values in arbritrary measurable space the proposition is no longer true, as pointed out in Bart Michels answer. – Ramiro Nov 12 '20 at 02:21

1 Answers1

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If you only ask that there is some sigma algebra $(Y, \mathcal N)$ such that "for all $f , g : X \to Y$ such that $f$ measurable and $f = g$ a.e., we have $g$ measurable", in general you can't conclude that the measure on $X$ is complete. For example, when $Y$ is a singleton or another set with the coarse sigma-algebra, any $f, g : X \to Y$ are measurable, and the assumption gives no information.

As soon as $Y$ contains a nontrivial measurable set, the implication is true. The argument when $Y = \mathbb R$ still works: let $\mathbf 0$ be any nontrivial measurable subset of $Y$ with complement $\mathbf 1$, and take $0 \in \mathbf 0$ and $1 \in \mathbf 1$. When $A \subset X$ is contained in a set of measure $0$, take $f$ to be the constant function $0$ and $g$ to be $1$ on $A$ and $0$ on $X - A$. Then $f$ is constant, hence measurable, and $f = g$ a.e. so that $g^{-1}(\mathbf 1) = A$ is measurable.

Bart Michels
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