# $f$ measurable, $f=g$ almost everywhere, complete measure space [duplicate]

Let $X$ be a nonempty set, $\mathcal{X}$ a $\sigma$-algebra of subsets of $X$, and $\mu$ a measure on $\mathcal{X}$ (i.e., $\mu:X\to[0,+\infty]$, $\mu(\phi)=0$, and $\mu$ is countably additive.

Proposition: If $f:X\to R$ is a measurable function, $f=g$ almost everywhere, then $g$ is a measurable function.

Can this proposition be proved in the the measure space is not complete ($\mathcal{X}$ contains all subsets of measure zero)? If not, can someone provide a counterexample?

Thanks.

• This is in the second case of the only answer in the linked post. – awllower Jul 9 '13 at 3:38

Assume that $f=0$ and $g$ is the characteristic function of a non-measurable set contained in a set of measure zero.