Exercise 3, § 3

Integration, V, p. 96

Let T, T' be two locally compact spaces, $\mu$ a positive measure on T, and $\mu'$ a positive measure on T'. On the product space $X = T \times T'$ , set $\lambda_t = \varepsilon_t \otimes \mu'$ , $\lambda'_{t'} = \mu \otimes \varepsilon_{t'}$ for $t \in T, t' \in T'$ . The mapping $t \mapsto \lambda_t$ (resp. $t' \mapsto \lambda'_{t'}$ ) of T (resp. T') into $\mathscr{M}(X)$ is vaguely continuous and $\mu$ -adequate (resp. $\mu'$ -adequate), and

\nu = \mu \otimes \mu' = \int \lambda_t d\mu(t) = \int \lambda'_{t'} d\mu'(t')

(Ch. III, §4, No. 1; cf. §8). Take for T the interval [0, 1] of $\mathbf{R}$ , for $\mu$ Lebesgue measure, for T' the interval [0, 1] equipped with the discrete topology, and for $\mu'$ the measure on T' defined by the mass +1 at every point of T'. Let f be the characteristic function of the 'diagonal' of X (the set of points (t, t), were t runs over [0, 1]). Show that f is upper semi-continuous and that

1 = \int^* d\mu(t) \int^* f(x) d\lambda_t(x) < \int^* f(x) d\nu(x) = +\infty

and

0 = \int^* f(x) d\nu(x) < \int^* d\mu(t) \int^* f(x) d\lambda_t(x) = 1 .

Solution

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