Let's call $\mu_{n}$ the Lebesgue measure on $\mathbb R^n$. Prove that if $A \subset \mathbb R^n$ and $B\subset \mathbb R^m$ are open subsets then $$ \mu_{n+m}(A\times B) = \mu_n(A)\mu_m(B). $$

In other words, this tells us that the Lebesgue $(n+m)$-dimensional measure and the product measure of $\mu_n$ and $\mu_m$ agree on open sets. How can I do? Suppose that $(A_i)_{i \in \mathbb N}$ and $(B_i)_{i \in \mathbb N}$ are sequences s.t. $$ A \subseteq \bigcup_{i} A_i, \qquad B \subseteq \bigcup_{j} B_j $$ Then $$ A \times B \subseteq \bigcup_{i} A_i \times\bigcup_{j} B_j $$ Now may I write - by $\sigma$-subadditivity $$ \mu_{n+m}(A \times B) \le \sum_{i,j} \mu_n(A_i)\mu_m(B_j) $$ Is it correct? Now I would like to conclude - by taking $\inf$ on RHS - that $\mu_{n+m}(A \times B) \le \mu_n(A)\mu_m(B)$: how can I do? And what about the opposite inequality?

Bonus (self-posed) question: what happens if we remove the hypothesis $A,B$ are open? For example, is it true that if $\mu_n(A)<\infty$ and $\mu_m(B)<\infty$ then $$ \mu_{n+m}^{\star}(A \times B) = \mu_n(A)\mu_m(B) $$ where $\mu^{\star}_{n+m}$ is the outer $(n+m)$-dimensional Lebesgue measure? Thanks in advance.


I suppose you want $(A_i)_i, (B_i)_i$ to be sequences of rectangular boxes. Taking the infimum over such sequences is ok and gives you "$\leq$".

But I think you prove the statement more easily with the uniqueness theorem for measures: For a rectangular box $R \in \mathbb{R}^m$ set $\alpha_R (A) = \mu_{n+m}(A \times R)$, $\beta_R (A) = \mu_n(A) \mu_m(R)$ (with the convention $0 \cdot \infty = 0$). Use the uniqueness theorem for measures to prove that $\alpha_R, \beta_R$ are coinciding measures on $\mathcal{B}(\mathbb{R}^n)$. Use it again to prove that $\mu_{n+m}(A \times \cdot)$ and $\mu_n(A) \mu_m(\cdot)$ coincide on $\mathcal{B}(\mathbb{R}^m)$ for all $A \in \mathcal{B}(\mathbb{R}^m)$.


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.