Exercise 1.2.22 (Lebesgue measure and Cartesian products)

We verify the definitions. Recall that we have already proven this theorem for elementary sets in Exercise 1.1.4 (Elementary measure of a product is product of elementary measures). It is not hard to generalise this result to open sets as well. In other words, if $U$ and $V$ are open sets, then $U\times V$ is an open, Lebesgue measurable set. By Lemma 1.2.11 we can represent both sets as unions of (almost) disjoint boxes $U={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }{B}_i$ with $m(U)={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }|B_i|$ and $V={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }{C}_i$ with $m(V)={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }|C_i|$. By the additivity of Lebesgue measure we then have:

(i)

Showing that ${(m^{d+d^{\prime }})}^{\text{∗}}(E\times F)\le {(m^d)}^{\text{∗}}(E)\times {(m^{d^{\prime }})}^{\text{∗}}(F)$ means $$\inf <!--\; FUNCTION\; APPLICATION\; -->\left\{\underset{n=1}{\overset{\infty }{\sum }}|B_n|:\begin{array}{c}{B}_n\text{ are boxes}\hfill \\ E\times F\subseteq \underset{n=1}{\overset{\infty }{\bigcup }}{B}_n\hfill \end{array}\right\}\le \inf <!--\; FUNCTION\; APPLICATION\; -->\left\{\underset{n=1}{\overset{\infty }{\sum }}|C_n|:\begin{array}{c}{C}_n\text{ are boxes}\hfill \\ E\subseteq \underset{n=1}{\overset{\infty }{\bigcup }}{C}_n\hfill \end{array}\right\}\times \inf <!--\; FUNCTION\; APPLICATION\; -->\left\{\underset{n=1}{\overset{\infty }{\sum }}|D_n|:\begin{array}{c}{D}_n\text{ are boxes}\hfill \\ F\subseteq \underset{n=1}{\overset{\infty }{\bigcup }}{F}_n\hfill \end{array}\right\}.$$

Pick an arbitrary box cover ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{C}_n$ of $E$ and ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{D}_n$ of $F$. Notice that the set ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{C}_n\times {\bigcup }_{n=1}^{\infty }{D}_n={\bigcup }_{n=1}^{\infty }{\bigcup }_{m=1}^{\infty }{C}_n\times D_m$ is a cover for $E\times F$, and a Cartesian product of two boxes is again a box. We thus obtain

$$\underset{(n,m)\in {\mathbb{N}}^2}{\sum }|C_n\times D_m|\le \underset{n=1}{\overset{\infty }{\sum }}|C_n|\times \underset{m=1}{\overset{\infty }{\sum }}|D_m|$$

(ii)

In the case where one of the quantities is infinite, the statement follows trivially. Thus, assume that both outer measures are finite. Let $U_E$ be an open cover of $E$ and let $U_F$ be an open cover of $F$ such that $m(U_E∕E)\le 𝜖$ and $m(U_F∕F)\le 𝜖$. Recall that a Cartesian product of two open sets $U_E\times U_F$ is again open with $E\times F\subseteq U_E\times U_F$. We now look at $(U_E\times U_F)∕(E\times F)$. Recall the disjointisation identity $(U_E\times U_F)∕(E\times F)=[U_E\times (U_F∕F)]\cup [(U_E∕E)\times F]$. By the subadditivity we get $$m^{\text{∗}}((U_E\times U_F)∕(E\times F))\le m^{\text{∗}}(U_E\times (U_F∕F))+m^{\text{∗}}((U_E∕E)\times F).$$

We demonstrate that both quantities can get arbitrarily small. Pick an $𝜖>0$. The trick is to combine covers the box covers of $U_E$ and $U_F∕F$, similarly for the the second term.

• Pick an arbitrary finite box cover ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{B}_n\supseteq U_E$. For the second part, by the measurability assumption we can find a box cover $U_F∕F\subseteq {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{K}_n$ such that ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }|K_n|\le \frac{𝜖}{2}∕{\sum }_{n=1}^{\infty }|B_n|$. The key observation is that the set ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{B}_n\times {\bigcup }_{n=1}^{\infty }{K}_n={\bigcup }_{(n,m)\in {\mathbb{N}}^2}{B}_n\times K_n$ covers $U_E\times (U_F∕F)$. We see that

  $$\underset{(n,m)\in {\mathbb{N}}^2}{\sum }|B_n|\times |K_n|\le \underset{n=1}{\overset{\infty }{\sum }}|B_n|\times \underset{n=1}{\overset{\infty }{\sum }}|K_n|\le \frac{𝜖}{2}$$

• Pick an arbitrary finite box cover ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{C}_n\supset F$. We find $U_F∕F\subseteq {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{L}_n$ such that ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }|L_n|\le \frac{𝜖}{2}∕{\sum }_{n=1}^{\infty }|B_n|$. Using the same logic as before we get

  $$\underset{(n,m)\in {\mathbb{N}}^2}{\sum }|C_n|\times |L_n|\le \underset{n=1}{\overset{\infty }{\sum }}|C_n|\times \underset{n=1}{\overset{\infty }{\sum }}|L_n|\le \frac{𝜖}{2}$$

Combining both approximations we see that $m^{\text{∗}}((U_E\times U_F)∕(E\times F))\le 𝜖$. Thus, $E\times F$ must be Lebesgue measurable, and we have

$$\begin{array}{llll}\hfill m(E\times F)& =m(U_E\times U_F)-m((U_E\times U_F)∕(E\times F))& \hfill & \\ \hfill & \ge m(U_E\times U_F)-𝜖=m(U_E)m(U_F)-𝜖& \hfill & \\ \hfill & =(m(E)-𝜖)(m(F)-𝜖)-𝜖& \hfill & \\ \hfill & =m(E)m(F)-𝜖\left[m(E)+m(F)+1\right]& \hfill & \end{array}$$

Adjusting $𝜖$ and taking limits $𝜖\to 0$ we see that $m(E\times F)\ge m(E)\times m(F)$. Combined with the first part of the exercise we obtain the desired result.