Suppose $(X,\mathcal{M},\mu)$ is a measure space, and $(X,\mathcal{M}^*,\mu^*)$ is its smallest compeletion, i.e.,
- $\mathcal{M}\subset\mathcal{M}^*$;
- $\mu^*|_{\mathcal{M}}=\mu$;
- any $\mu$-null set belongs to $\mathcal{M}^*$.
where $\mu$-null set is a subset of $X$ that can be contained within a set of measure zero with respect to $(\mathcal{M},\mu)$.It's easy to show that the exact expresion of this extension is:
$$\begin{align}\mathcal{M}^*&=\{A\cup S:A\in\mathcal{M},S\text{ is a }\mu\text{-null set}\},\\[2ex]\mathcal{M}^*&\stackrel{\mu^*}{\longrightarrow}[0,\infty],\quad A\cup S\mapsto\mu(A).\end{align}$$
Given an extended real-valued function $f:D_f(\subset X)\to[-\infty,\infty]$, if there exists a function $g:D_g\to[-\infty,\infty]$ s.t.
- $D_f\Delta D_g$ is $\mu$-null,
- $\{x\in D_f\cap D_g:f(x)\ne g(x)\}$ is $\mu$-null,
then we say $f=g$ a.e. (almost everywhere). If $D_g=X$ and $g$ is measurable w.r.t $(\mathcal{M},\mu)$, we say $f$ is $\mu$-a.e. measurable.
Let $\tilde{f}$ be any extended function of $f$ on $X$, my question is how to prove that
$$''f\text{ is a.e. }\mu\text{-null measurable} \Longleftrightarrow \tilde{f}\text{ is measurable w.r.t. } (\mathcal{M}^*,\mu^*)''$$