Exercise 1.4.18 (Borel measurable sets and Cartesian products II)

First notice that the statement holds for closed sets $F\subseteq P({ℝ}^{d_1+d_2})$. Let $E\in F$. Then any convergent sequence ${(x_n^{(1)},\dots ,x_n^{(d_1)},x_n^{(d_1+1)},\dots ,x_n^{(d_1+d_2)})}_{n\in \mathbb{N}}$ converges in $E$. In particular, its ${(x_n^{(d_1+1)},\dots ,x_n^{(d_1+d_2)})}_{n\in \mathbb{N}}$ part converges in $\{x_2\in {ℝ}^{d_2}:(x_1,x_2)\in E\}$. But any convergent sequence in $\{x_2\in {ℝ}^{d_2}:(x_1,x_2)\in E\}$ can be expressed as a part of a convergent sequence in $E$; thus, $\{x_2\in {ℝ}^{d_2}:(x_1,x_2)\in E\}$ must be closed too. In particular, it is Borel measurable. Thus, $P(E)$ holds for any $E\in F$. Following Remark 1.4.15 we now demonstrate that $P$ is a $\sigma$-property.

(i)

For $\varnothing$ we have $\{x_2\in {ℝ}^{d_2}:(x_1,x_2)\in \varnothing \}=\varnothing$ is Borel measurable.

(ii)

Suppose that $P(E)$ is true, i.e., $\{x_2\in {ℝ}^{d_2}:(x_1,x_2)\in E\}$ is Borel measurable. We then have $$\left\{x_2\in {ℝ}^{d_2}:(x_1,x_2)\in {ℝ}^{d_1+d_2}∕E\right\}={ℝ}^{d_2}∕\{x_2\in {ℝ}^{d_2}:(x_1,x_2)\in E\}$$

and the latter is Borel measurable by axiom 2.

(iii)

Suppose that ${\left(E\right)}_{n\in \mathbb{N}}$ is such that $P(E_n)$ is true. We must demonstrate that $P({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{E}_n)$ is true. $$\left\{x_2\in {ℝ}^{d_2}:(x_1,x_2)\in \bigcup _{n=1}^{\infty }{E}_n\right\}=\bigcup _{n=1}^{\infty }\{x_2\in {ℝ}^{d_2}:(x_1,x_2)\in E\}.$$

This must be Borel measurable by the closure under countable unions property.

Thus, $P(E)$ holds for all $E\in ⟨F⟩=B$.

Now recall from Lemma 1.2.13 that every null set is Lebesgue measurable. In particular, $\{0\}\times V\subset {ℝ}^2$, where $V$ is the Vitali set, is a null set. Thus, although the product $\{0\}\times V\subset {ℝ}^2$ is Lebesgue measurable, any slice of this set is not Lebesgue measurable in $ℝ$.