Suppose that is $f$ is a measurable function, and $g$ differs with $f$ only by a measure zero set, is $g$ measurable?

Question

Suppose that is $f$ is a measurable function, and $g$ differs with $f$ only by a measure zero set, is $g$ measurable? I do not think this is necessarily right, but is there any counter example?

Answer 1

Hint:

1. The sum of two measurable functions is measurable.
2. For any $E$, either $\{x: f(x)-g(x) \in E\}$ or its complement has measure $0$.

Answer 2

This depends on whether your measure is complete. Consider, for example, Lebesgue measure on the Borel subsets of the unit interval $[0,1]$. Let $f(x)=x$ for $x\in[0,1]$. Let $K\subset[0,1]$ denote the Cantor set. Choose a non-Borel-measurable subset $B$ of $K$. Define $g(x)=f(x)$ for $x\in [0,1]\setminus K$ and $g(x) =2+1_B(x)$ for $x\in K$. Then $\{x:f(x)\not=g(x)\}=K$ is a measurable set of measure $0$, but $g^{-1}(\{3\})=B$, so $g$ is not (Borel) measurable.