Exercise 13, § 5

Let T be a locally compact space, $\mu$ a positive measure on T, and $(f_n)$ a sequence of functions in $\mathscr{L}^1(T, \mu)$; set $\mu_n = f_n \cdot \mu$ and, for every $\mu$-measurable set A,

a) Show that if T is not compact, it can happen that the sequence $(f_n)$ is not bounded in $\mathscr{L}^1(T, \mu)$ but the sequence of integrals $\int f_n g \, d\mu$ is bounded for every $g \in \mathscr{H}(T)$.

b) Show that if the sequence $(\mu_n(A))$ is bounded for every subset A of T reduced to a point and for every open set A such that the measure induced by $\mu$ on the boundary of A has finite support, then the sequence $(f_n)$ is bounded in $\mathscr{L}^1$, or, what amounts to the same, the sequence of norms $(\| \mu_n \|)$ is bounded. (Show first that every point $t_0$ of T admits an open neighborhood U such that the sequence of numbers $|\mu_n|(U)$ is bounded. For this, one argues by contradiction by proving that, in the contrary case, one could construct a strictly increasing sequence $(n_k)$ of integers, a decreasing sequence $(U_k)$ of neighborhoods of $t_0$, and a sequence $(W_k)$ of open sets quadrable for $\mu$ (Ch. IV, §5, Exer. 17) having the following properties: $\overline{U}_k \subset U_{k-1}$, $\mu(U_k - \{t_0\}) \leq 1/k$, $|\mu_{n_i}|(U_k - \{t_0\}) \leq 1$ for $i < k$, $\overline{W}_k \subset U_k - \overline{U}_{k+1}$, and finally

Finally, consider the union W of the $W_k$, to obtain a contradiction (the 'gliding hump method'). Then show by an analogous argument that there exists a compact subset K of T such that the sequence $|\mu_n|(T - K)$ is bounded.)

c) Deduce from b) that if, for every open set A of T, the sequence $(\mu_n(A))$ is bounded, then the sequence $(f_n)$ is bounded in $\mathscr{L}^1$.

d) On the interval $[0, 1]$, if one takes for $\mu_n$ the measure defined by the mass n at the point $t = 0$ and the mass $-n$ at the point $t = 1/n$, then the sequence $(\mu_n(A))$ is bounded for every open set A that is quadrable for all the measures $|\mu_n|$; however, the sequence of norms $\| \mu_n \|$ is not bounded. Similarly, if one takes for $\mu_n$ the measure defined by the mass n at the point $t = 1/n$ and the mass $-n$ at the point $t = 1/(n+1)$, then the sequence $(\mu_n(A))$ is bounded for every interval A contained in $[0, 1]$ (hence for every finite subset A) without the sequence $(\| \mu_n \|)$ being bounded.