Changing a r.v. on a set of measure zero leaves it measurable

Question

This is a claim that I thought and wanted to ask for a check.

Claim. Take $X$ a real r.v. on a measure space $(\Omega,\Sigma,P)$, with $\Sigma$ complete. Take a set of measure zero $U$ and define $Y$ s.t.:

• $Y(\omega)=X(\omega)$ if $\omega \in U^{c}$
• $Y(\omega)$ arbitrary if $\omega \in U$

Than $Y$ is also a r.v..

Proof. We just have to check that $Y$ is measurable. To do this we fix $B$ a borel set and check if $Y^{-1}(B) \in \Sigma$. We note that :

$Y^{-1}(B)=\{ Y^{-1}(B) \cap U\} \cup \{ Y^{-1}(B) \cap U^{c}\}=\{ Y^{-1}(B) \cap U\} \cup \{ X^{-1}(B) \cap U^{c}\}$

Now:

• $X^{-1}(B)$ is measurable
• $U$ is measurable, otherwise in the claim we could not even say Take a set of measure zero $U$. Therefore also $U^c$ is measurable.
• $\{ Y^{-1}(B) \cap U\} \subset$ U and since the sigma algebra is complete this is measurable.

Therefore $Y$ is measurable.

Is my proof/claim correct, and is the completeness condition necessary ?

Answer 1

Yes your proof is correct.

And yes the completeness is necessary. Otherwise there exists a measure zero set $U\subset\Omega$ which contains a non-measurable set $A$. Then $1_A$ is equal to the random variable $0$ on $A^\complement$, but is not itself a random variable, because $A$ is not measurable.