Is the set of two measurable functions being equal to each other a measurable set?

Question

Let $(\Omega,F,\mu)$ be a measure space that is not necessarily complete. Let's assume the measure $\mu$ is finite. Let $f$ and $g$ be measurable functions such that $f=g$ almost everywhere. Is it true to state that $\{f=g\}$ is in $F$?

Answer 1

I assume that you consider functions $\Omega \to \mathbb R$. Since you assumed that $f$ and $g$ are measurable, then $f-g$ is measurable. It means that the set $\{f-g = 0\} = \{f-g \ge 0\} \cap \{f-g \le 0\}$ is measurable. So you don't even need "$f=g$ almost everywhere" condition.

On the other hand, if $f$ and $g$ are not measurable, then "$f=g$ almost everywhere" only means that $\{f \ne g\} \subseteq N$, where $\mu(N) = 0$. Since you didn't assume that your space is complete, you can't say that $\{f \ne g\}$ is measurable, and hence can't say that $\{f=g\}$ is measurable. So "$f=g$ almost everywhere" doesn't help here either.