1
$\begingroup$

If we define the outer product measure like this : $$ \mu*\nu(E) =\inf\{\sum_{i=1}^\infty \mu(A_i)\nu(B_i)\mid E\subset\bigcup_{i=1}^\infty A_i\times B_i,\ \text{$A_i$ is $\mu$- measurable, $B_i$ is $\nu$-measurable}\}. $$ Is it true that if that whenever $E=A\times B\subset X\times Y$ is $\mu*\nu$-measurable, then $A\subset X$ is $\mu$-measurable and $B\subset Y$ is $\nu$-measurable? If it's not true, then what would the counterexample be? I don't know where to start.

The other way around it's easy to show, I was able to prove it, moreover $\mu*\nu(A\times B)=\mu(A)\nu(B)$ if $A$ and $B$ are $\mu$ and $\nu$-measurable respectively.

$\endgroup$
  • 1
    $\begingroup$ No. It is not true. Let $A$ be any NON-measurable subset of $X$ and let $B$ be any measurable subset of $Y$, such $\nu(B)=0$. Then $A\times B$ is $\mu*\nu$-measurable. $\endgroup$ – Ramiro Sep 29 '15 at 15:18
1
$\begingroup$

No. It is not true. Let $A$ be any NON-measurable subset of $X$ and let $B$ be any measurable subset of $Y$, such $\nu(B)=0$. Then $A \times B$ is $\mu*\nu$-measurable.

More details: Let us prove that $A \times B$ is $\mu*\nu$-measurable. Let $\mu^*(A)$ be the outer measure of $A$. Then, $\mu*\nu(A \times B)$ the outer product measure of $A \times B$, satisfies: $$ \mu*\nu(A \times B)= \mu^*(A).\nu(B)=0 $$ Since $\mu*\nu(A \times B)=0$, we have that $A \times B$ is $\mu*\nu$-measurable.

$\endgroup$
  • $\begingroup$ I don't really follow why such cartesian product would be measurable. Could you explain with a bit more details? $\endgroup$ – Proton Sep 29 '15 at 16:35
  • $\begingroup$ @Proton I have added a more detailed explanation to the answer. Please, let me know if you have any further question. $\endgroup$ – Ramiro Sep 29 '15 at 20:25
  • $\begingroup$ Ah right, zero measure sets would indeed be measurable, I didn't notice this fact. Silly me :) Now everything is clear. Thank you! $\endgroup$ – Proton Sep 29 '15 at 20:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.