 abbracciopj

2022-06-29

While studying concepts of measurable functions, there was a theorem suggesting that if any measurable function is altered on a null set, its measurability still remains.
If $f:E\to \mathbb{R}$ is measurable, $E\in \mathcal{M},g:E\to \mathbb{R}$ is such that the set $\left\{x:f\left(x\right)\ne g\left(x\right)\right\}$ is null, then $g$ is measurable.
They showed the following proof.
Consider the difference $d\left(x\right)=g\left(x\right)-f\left(x\right)$. It is zero except on a null set, so
$\left\{x:d\left(x\right)>a\right\}=\left\{\begin{array}{ll}\text{a null set}& a\ge 0\\ \text{a full set}& a<0\end{array}$
Here, a full set is the complement of a null set. Since both null and full sets are measurable, $d$ is a measurable function. $g=f+d$ is thus measurable.
Now, I was curious whether the statement remains true if I change the set $\left\{x:f\left(x\right)\ne g\left(x\right)\right\}$ into $\left\{x:f\left(x\right)=g\left(x\right)\right\}$ , i.e., differ at points in a full set, because there was no doubt if I alter the proof as the following.
The difference $d\left(x\right)$ is still measurable since
$\left\{x:d\left(x\right)>a\right\}=\left\{\begin{array}{ll}\text{a full set}& a\ge 0\\ \text{a full set}& a<0\end{array}$
However, the later statement is actually not plausible at all. Is there any contradictory logic among here?
Appreicate as always. robegarj

Expert

Let $f=0,A\subseteq E$ be a non-measurable subset and $N$ is a null set with $N\cap A=\varnothing .$
Then we let $g\left(x\right)=\left\{\begin{array}{ll}1,& x\in A\\ 0,& x\in N\\ -1,& x\in E\setminus \left(A\cup N\right)\end{array}$
Thus,$\left\{x:d\left(x\right)>0\right\}=A$ which is non-measurable. So here $d\left(x\right)$ is not a measurable function.

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