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Let $K$ be a compact Hausdorff space. Denote by $ca_r(K)$ the set of all countably additive, signed Borel measures which are regular and of bounded variation. Let $(\mu_n)_{n\in\mathbb{N}}\subset ca_r(K)$ be a bounded sequence satisfying $\mu_n\geq 0$ for all $n\in\mathbb{N}$. Let $f:K\rightarrow\mathbb{R}^n$ be continuous and bounded and assume that the limit $z:=\lim_{n\rightarrow\infty}\int fd\mu_n\in\mathbb{R}^n$ exists. Is there a $\mu\in ca_r(K)$, $\mu\geq 0$, such that $z=\int fd\mu$? This question is related to this.

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Andy Teich
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2 Answers2

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Yes. Let the limit be $L$. Let $\min \{f(x): x \in K\} = f(a)$, $\max \{f(x): x \in K\} = f(b)$. Now $ \|\mu_n\| f(a) \le \int f \ d\mu_n \le \|\mu_n\| f(b)$, so there is $\lambda_n \in [0,1]$ such that $\int f \ d\mu_n = \|\mu_n\| (\lambda_n f(a) + (1-\lambda_n) f(b)$. Taking a subsequence (and noting that $\|\mu_n\|$ is bounded), we can assume $\|\mu_n\|$ and $\lambda_n$ converge, and then $\mu = \lim_{n} \|\mu_n\| (\lambda_n \delta(a) + (1-\lambda_n) \delta(b)$ (where $\delta(a)$ and $\delta(b)$ are the unit point masses at $a$ and $b$) satisfies the requirements.

Robert Israel
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  • Thanks for the very nice answer! I realized what the problems are and therefore changed the question and wrote down the full question I had in mind... – Andy Teich Sep 19 '12 at 15:50
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It suffices to handle multidimensional $f$. To simultaneously handle $f$ and $g$ you could just look at the function $h=(f,g)\in\mathbb{R}^{n+1}$.

Taking a subsequence, we may assume that $\mu_n(K)\to m$ for some $0\leq m<\infty$. If $m=0$, we let $\mu=0$ otherwise we use the following argument.

Assume that $m>0$. Define $\nu_n=\mu_n/\mu_n(K)$ so that $\nu_n$ are probability measures. For every $n$, the value of $\int f(x)\,\nu_n(dx)$ belongs to the convex hull of $f(K)$, which is a compact subset of $\mathbb{R}^n$. The limit $z/m$ also belongs to this set. That is, we can write $z/m=\sum_{i=1}^N \lambda_i\, f(k_i)$ for some weights $\lambda_i>0$ with $\sum_{i=1}^N \lambda_i=1$.

Therefore $z=\int f(x)\,\mu(dx)$ where $\mu=m\sum_{i=1}^N \lambda_i \delta_{k_i}.$

This is just a modification of Robert Israel's answer.

  • What is if we drop boundedness of $f$? I realized that in my case, $f$ is only bounded in one component, and not in the others... – Andy Teich Sep 20 '12 at 08:25
  • @AndyTeich So $f$ is not continuous? Then the result is false, even for bounded $f$. –  Sep 20 '12 at 12:06
  • I meant the case when $f$ is unbounded but continuous. But then $f(K)$ is still compact, which I had overseen... – Andy Teich Sep 20 '12 at 14:12