Let $X\subseteq \mathbb{R}$ and $f$ a measurable function. Suppose that I find: $f=0$ a.e. on $X$ and $f\neq 0$ a.e. on $X$. I believe that it is not a contradiction itself if $\mu(X)=0$. On the other hand, if $\mu(X)>0$, then $$X=(X\cap\{f=0\})\cup (X\cap \{f\neq 0\})$$ implying $\mu (X)\leq 0,$ using the subadditivity of the measure. Indeed, this is a contradiction.
Are these ideas correct?
For example, if $X=\emptyset$, then $f=0$ a.e. on $X$ and $f\neq 0$ a.e. on $X$ would be vacuously true.