 Solution

The relation is reflexive because the exceptional set for $f=f$ is empty, and symmetric because the exceptional sets for $f=g$ and $g=f$ agree. If $f=g$ off a null set $N_1$ and $g=h$ off a null set $N_2$, then $f=h$ off $N_1\cup N_2$, which is null by part (b). Hence it is transitive as well, so equality almost everywhere is an equivalence relation.