# Are measurable under product outer measure sets a cartesian product of measurable sets?

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.

• 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. – Ramiro Sep 29 '15 at 15:18

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.

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