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Let $X$ be a compact metric space and $\mu_n$ a sequence of finite Borel measures on $X$ with the property that

$$\sup_n \mu_n(X)<\infty.$$

Show for all $f \in C(X)$ there exists a subsequence $n_j$ such that $\int f d \mu_{n_j}$ converges.

Attempt:

We know as $X$ is a compact metric space, that $C(X)$ is separable. I.e., there exists a countable dense subset, call it $A \subset C(X)$ and so

$$A:=\{f_n: n \in \Bbb{N}, \text{$f_n:X \to \Bbb{R}$ is continuous}\}.$$

and $\bar{A}=X$. I also know for each $n$, $f_n(X)$ attains a min and max. But I dont see how to produce this convergent subsequence. Also, since the sup of the $\mu_n(X)$ is finite and I have a sequence I cant say they have a convergent subsequence right cause they live in $\Bbb{R}^{\geq 0}$ which is not compact..

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1 Answer 1

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put $A=\sup_n \mu_n(X)$.Since $X$ is compact, $f$ is bounded, so there is some constants $m,M$ such that: $\forall x\in X:m\leq f(x) \leq M $,so $mA\leq \int f d \mu_{n} \leq MA$ \ This shows that $u_{n}:=\int f d \mu_{n}$ is a bounded sequence of real numbers,so it has a convergent subsequence $\int f d \mu_{n_{j}}$

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