Here I give the definition of an almost everywhere measurable function:
Given $(X, \mathcal{F}, \mu)$ a measurable space. A function $f: X \to \mathbf{R}$ is $\mu$-almost everywhere measurable if there exists $N \in \mathcal{F}$ with $\mu(N) = 0$ such that for all $\alpha \in \mathbf{R}$ that $f^{-1}((\alpha, +\infty)) \cap (X – N) \in \mathcal{F}$.
I would think that if a function $f$ is almost everywhere measurable, then $f$ must be equal to a measurable function $g$ almost everywhere. However, I am starting to think this might not be the case. My attempt is to note $f = f\chi_{X – N}$ almost everywhere and show $f\chi_{X – N}$ is measurable by definition: We let $\alpha \in \mathbf{R}$. Then $f\chi_{X – N}^{-1}((\alpha, +\infty)) = \begin{cases}
f^{-1}(\alpha, +\infty) \cap (X – N) &, \mbox{ if } \alpha \geq 0; \\
f^{-1}(\alpha, + \infty) &, \mbox{ otherwise}.
\end{cases}$
However, it is not necessarily true that $f^{-1}(\alpha, +\infty) \in \mathcal{F}$ as $f$ is not necessarily measurable. I couldn't figure out how to fix this. Is the claim that I am trying to show here false entirely?
Update:
I just realized that in the case when $\alpha < 0$, we really have $(f^{-1}(\alpha, +\infty) \cap (X – N)) \cup N$, which would be measurable. Is this correct?
Best Answer
The issue was an error in computing the preimage (which I also initially made) - $f\cdot\chi_{X\setminus N}$ is actually measurable. For any $(\alpha,\infty)\subseteq\Bbb R$, $(f\cdot\chi_{X\setminus N})^{-1}(\alpha,\infty)$ is precisely $f^{-1}(\alpha,\infty)\cap(X\setminus N)$ and this is measurable by hypothesis on $f$ and $N$, if $\alpha\ge 0$, and if $\alpha\lt0$ - as you rightly point out - the preimage is not just $f^{-1}(\alpha,\infty)\cap(X\setminus N)$. It is $N\sqcup f^{-1}(\alpha,\infty)\cap(X\setminus N)$, but this is a union of measurable sets which is measurable.
Since intervals of the form $(\alpha,\infty)$ generate the Borel algebra, $f\cdot\chi_{X\setminus N}$ is a measurable function in the usual sense, and $f\cdot\chi_{X\setminus N}=f$ almost everywhere since $N$ is a null set.
The claim is correct.