I am self studying measure theory.I measure theory it is often important to check if a function is measurable.If the function is continuous then it is measurable of course.But if the function is not continuous,then we have no other option than checking each $f^{-1}(a,\infty)$ to be measurable.So,I was looking for a sufficient condition for a function to be measurable.Suddenly the following came into my mind:
Let $(X,\mathcal S,\mu)$ be a measure space and $f:X\to \mathbb R$ be a measurable function.Suppose $g:X\to \mathbb R$ is a function such that $\mu\{x\in X:f(x)\neq g(x)\}=0$,then $g$ is measurable .
I am not sure whether the statement is true or not.So,can someone help me?
Not true without the additional assumption of completeness of the space.
Let $X=\mathbb R, \mathcal S =$ Borel sigma algebra and $\mu$ be Lebesgue measure. Let $C$ be the Cantor set. There exists a non-Borel set $E$ contained in $C$. Let $f=\chi_{C^{c}}$ and $g=\chi_{E\cup C^{c}}$. Then $f=g$ a.e. $\mu$ but $\{x: g(x)=1\}$ is not a Borel set.
Notations: $\chi_A(x)=1$ if $x \in A$ and $0$ otherwise. $A^{c}$ denotes $\mathbb R \setminus A$.
If the measure space is complete then $\{x: g(x)<a\}$ and $\{x: f(x)<a\}$ differ only by set of measure $0$ so the mesurability of the second set implies that of the first. To be explicit @KishalaySarkar $\{x: g(x)<a\}= [\{x: f(x)<a\} \setminus \{x:f(x)\neq g(x)\} ]\cup [\{x: g(x)<a\} \cap \{x: f(x)\neq g(x)\}]$.