7
$\begingroup$

I am stuck on this problem from Folland's Real Analysis, Second Edition:

For $j = 1, 2$, let $\mu_j, \nu_j$ be $\sigma$-finite measures on $(X_j, \mathcal{M}_j)$ such that $\nu_j <\!\!< \mu_j$. Then $\nu_1 \times \nu_2 <\!\!< \mu_1 \times \mu_2$.

Here is about where I am at.

(1) It is immediate that if $\mu_1 \times \mu_2(A \times B) = 0$, then $\nu_1 \times \nu_2(A \times B) = 0$.

(2) Because $\nu_1 \times \nu_2$ is $\sigma$-finite, it is enough to prove the result for when $\nu_1 \times \nu_2$ is finite. Since the result is quickly verified if either $\nu_1$ or $\nu_2$ is the zero measure, we may assume that both $\nu_1$ and $\nu_2$ are finite. We then have the following equivalent formulation for $\nu_1 <\!\!< \mu_1$:

For every $\epsilon > 0$, there exists $\delta > 0$ such that $\mu_1(E) < \delta$ implies $\nu_1(E) < \epsilon$.

And similarly for $\nu_2 <\!\!< \mu_2$.

I thought I might be able to prove the same condition for $\nu_1 \times \nu_2$ with respect to $\mu_1 \times \mu_2$, which would then imply $\nu_1 \times \nu_2 <\!\!< \mu_1 \times \mu_2$ since $\nu_1 \times \nu_2$ has been reduced to being finite. As a suggestion, following Folland's technique, it might be easier to argue by contradiction, assuming first the $\epsilon-\delta$ condition is false.

(3) If $\mu_1 \times \mu_2(E) = 0$, then, by definition,

$$0 = \inf \bigg\{ \sum_n \mu_1(A_n)\mu_2(B_n) \colon A_n \times B_n \text{ are rectangles such that } E \subset \bigcup_n A_n \times B_n\bigg\}.$$

To show $\nu_1 \times \nu_2(E) = 0$, we want to show that the analogous equation holds for $\nu_1 \times \nu_2$.

I can't seem to put the pieces together, despite some effort.

Any help would be greatly appreciated. Thanks.

$\endgroup$
4
$\begingroup$

Assume that $E \in M_1 \times M_2$ such that

$\mu_1\times \mu_2(E)=0$. We have to show that $\nu_1\times \nu_2(E)=0$.

By Fubini theorem we have

$$0=\mu_1\times \mu_2(E)=\int_{X_1}\mu_2(E \cap (\{x\} \times X_2))d\mu_1(x).$$

This means that $$ \mu_{1}(\{ x : \mu_2(E \cap (\{x\} \times X_2))>0 \})=0, $$

equivalently

$$ \mu_{1}( X_1 \setminus \{ x : \mu_2(E \cap (\{x\} \times X_2))=0 \})=0. $$

Since $\nu_1 \ll \mu_1$ we have

$$ \{ x : \mu_2(E \cap (\{x\} \times X_2))=0 \} \subseteq \{ x : \nu_2(E \cap (\{x\} \times X_2))=0 \}. $$

Since $\nu_1 \ll \mu_1$ and

$$ \mu_{1}( X_1 \setminus \{ x : \nu_2(E \cap (\{x\} \times X_2))=0\})=0, $$ we have $$ \nu_{1}( X_1 \setminus \{ x : \nu_2(E \cap (\{x\} \times X_2))=0\})=\nu_{1}(\{ x : \nu_2(E\cap (\{x\} \times X_2))>0\})=0. $$ Finally, we get

$$\nu_1\times \nu_2(E)=\int_{X_1}\nu_2(E \cap (\{x\} \times X_2))d\nu_1(x)= $$ $$ \int_{\{ x : \nu_2(E \cap (\{x\} \times X_2))>0\} }\nu_2(E \cap (\{x\} \times X_2))d\nu_1(x)+ $$ $$ \int_{\{ x : \nu_2(E \cap (\{x\} \times X_2))=0\}}\nu_2(E \cap (\{x\} \times X_2))d\nu_1(x)=0. $$

$\endgroup$

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.