Almost everywhere equal implies measurable

Question

I am still new to Lebesgue measure theory. I would like to get a verification on my proof.

Statement: Assume a complete measure space. If $f$ is a measurable function and $f=g$ almost everywhere, then $g$ is measurable.

Proof: Let $c$ be a real number and $N$ the null set such that $f=g$ on $N^C$. Hence, $$g^{-1}(c,\infty)=[g^{-1}(c,\infty)\cap N]\cup[g^{-1}(c,\infty)\cap N^C].$$

The first RHS bracket is measurable since it is the subset of $N$ and the space is complete. The second is measurable because it is equal to $f^{-1}(c,\infty)\cap N^C$, where $f^{-1}(c,\infty)$ is measurable by assumption and $N^C$ is measurable because $N$ is. Q.E.D.

Are there missing details in the proof? Thanks a lot.

Answer 1

In general, if we consider a measurable function $f: (\Omega_0, \mathcal{F}_0) \to (\Omega_1, \mathcal{F}_1)$ between two measurable spaces, $f$ is called measurable iff $$\forall B \in \mathcal{F}_1\quad f^{-1}(B) = \{\omega \in \Omega_0 \mid f(\omega) \in B\} \in \mathcal{F_0}$$ Now we have $g = f$ a.e. on $\mu$ -- a measure defined on $(\Omega_1, \mathcal{F}_1)$. If we denote $A = \{\omega \in \Omega_0 \mid g(\omega) \neq f(\omega)\}$ as the set that they disagree on, we know that (1) $A \in \mathcal{F}_0$ and (2) $\mu(A) = 0$.

To show that $g$ is also measurable, we follow the definition and take any $B \in \mathcal{F}_1$, and note that: $$ \begin{align} g^{-1}(B) &= \{\omega \in \Omega_0 \mid g(\omega) \in B\} \\ &= \{\omega \in \Omega_0 \mid g(\omega) \in B, \omega \in A \} \cup \{\omega \in \Omega_0 \mid g(\omega) \in B, \omega \in A^c\} \end{align} $$ and our goal is to show $g^{-1}(B) \in \mathcal{F}_0$. For the first part, notice that this is a subset of $A$, which is a null set in $\mu$ (i.e. $\mu(A) = 0$). This is where the completeness of measure space makes things easy -- any subset of a null set must be inside $\mathcal{F}_0$ by definition.

Next, for the second part, we easily have $$ \{\omega \in \Omega_0 \mid g(\omega) \in B, \omega \in A^c\} = \{\omega \in \Omega_0 \mid f(\omega) \in B, \omega \in A^c\} $$ since $f = g$ on $A^c$, and the latter set is simply $f^{-1}(B) \cap A^c$. Now since $f$ is measurable we have $f^{-1}(B) \in \mathcal{F}_0$, and $A^c \in \mathcal{F}_0$ trivially as well. We thus show that $g^{-1}(B) \in \mathcal{F}_0$.