Exercise 2, § 8

Let T be the interval $[0,1]$ of $\mathbf{R}$, equipped with the topology induced by that of $\mathbf{R}$, and T' the interval $[0,1]$ of $\mathbf{R}$, equipped with the discrete topology; let $\mu$ be Lebesgue measure on T, $\mu'$ the discrete measure on T' defined by the mass +1 at every point of T', and let $\nu = \mu \otimes \mu'$ be the product measure on $X = T \times T'$.

a) For every $t \in T$, the set $\{t\}$ is $\mu$-negligible, but the set $\{t\} \times T'$ is not $\nu$-negligible.

b) The 'diagonal' $\Delta$ of $T \times T'$ (the set of points $(t,t)$, where t runs over $[0,1]$) is a closed set in $T \times T'$, locally $\nu$-negligible but not $\nu$-negligible, such that $\mu'(\Delta(t)) = 1$ for all $t \in T$ and $\mu(\Delta(t')) = 0$ for all $t' \in T'$ (cf. §3, Exer. 3).

c) On the space $X \times X = Y$, consider the product measure $\rho = \nu \times \nu$. Show that there exists in Y a set A, locally $\rho$-negligible, such that for every $x \in X$, $A(x)$ and $-1$ $A(x)$ are $\nu$-measurable and not locally negligible. (If $x = (t,t')$, where $t \in T, t' \in T'$, take A to be such that $A(x)$ contains all of the points $(s,t) \in X$, where $0 \leq s \leq 1$.)