abbracciopj

Answered

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 $\{x:f(x)\ne g(x)\}$ is null, then $g$ is measurable.

They showed the following proof.

Consider the difference $d(x)=g(x)-f(x)$. It is zero except on a null set, so

$\{x:d(x)>a\}=\{\begin{array}{ll}\text{a null set}& a\ge 0\\ \text{a full set}& a<0\end{array}$

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 $\{x:f(x)\ne g(x)\}$ into $\{x:f(x)=g(x)\}$ , i.e., differ at points in a full set, because there was no doubt if I alter the proof as the following.

The difference $d(x)$ is still measurable since

$\{x:d(x)>a\}=\{\begin{array}{ll}\text{a full set}& a\ge 0\\ \text{a full set}& a<0\end{array}$

However, the later statement is actually not plausible at all. Is there any contradictory logic among here?

Appreicate as always.

If $f:E\to \mathbb{R}$ is measurable, $E\in \mathcal{M},g:E\to \mathbb{R}$ is such that the set $\{x:f(x)\ne g(x)\}$ is null, then $g$ is measurable.

They showed the following proof.

Consider the difference $d(x)=g(x)-f(x)$. It is zero except on a null set, so

$\{x:d(x)>a\}=\{\begin{array}{ll}\text{a null set}& a\ge 0\\ \text{a full set}& a<0\end{array}$

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 $\{x:f(x)\ne g(x)\}$ into $\{x:f(x)=g(x)\}$ , i.e., differ at points in a full set, because there was no doubt if I alter the proof as the following.

The difference $d(x)$ is still measurable since

$\{x:d(x)>a\}=\{\begin{array}{ll}\text{a full set}& a\ge 0\\ \text{a full set}& a<0\end{array}$

However, the later statement is actually not plausible at all. Is there any contradictory logic among here?

Appreicate as always.

Answer & Explanation

robegarj

Expert

2022-06-30Added 24 answers

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 $g(x)=\{\begin{array}{ll}1,& x\in A\\ 0,& x\in N\\ -1,& x\in E\setminus (A\cup N)\end{array}$

Thus,$\{x:d(x)>0\}=A$ which is non-measurable. So here $d(x)$ is not a measurable function.

Then we let $g(x)=\{\begin{array}{ll}1,& x\in A\\ 0,& x\in N\\ -1,& x\in E\setminus (A\cup N)\end{array}$

Thus,$\{x:d(x)>0\}=A$ which is non-measurable. So here $d(x)$ is not a measurable function.

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