Let $\epsilon>0$ be given. Let $Y$ be a compact, Hausdorff space and let $U\subseteq Y$ be an open subset. Assume that $(\mu_n)_{n\in\mathbb{N}}$ is a sequence of regular Borel probability measures on $Y$. I want to show the existence of a continuous function $f$, supported on $U$, with $0\leq f\leq 1$ and s.t. $\mu_n(\{x\in U: f(x)\neq 1\})<\epsilon$ for all $n\in \mathbb{N}$.

What have I tried:

Clearly, we can not take just the characteristic function on $U$, since this function will not be continuous on $Y$.

By regularity of the measures, we can choose for every $n$, a compact subset $K_n\subseteq U$ s.t. the measure $\mu_n(K_n)>\mu(U)-\epsilon$. Now, define $K:=\bigcup_{n\in\mathbb{N}}K_n$. $K\subseteq U$ and it is not clear anymore that $K$ is closed! But if it was closed, by Urysohn's lemma there exists the function that takes the value $1$ on $K$ and $0$ outside $U$.

I don't know how to fix this idea.

Thanks for any help

  • $\begingroup$ As mentioned below it isn't true as stated, but it could be true with additional assumptions. For instance, if the sequence $\mu_n$ were tight in $U$, or were weakly converging to some measure supported in $U$. $\endgroup$ – Nate Eldredge Feb 5 '18 at 15:04

The statement you want to prove seems wrong: for instance on $K=[0,1]$ with $U=]0,1[$ let $(q_n)$ be countable dense in $U$, and let $\mu_n$ be the Dirac measure on $q_n$. Because the measures are Dirac, you are asking for a continuous function $f$ such that for each $n$, $f(q_n)=1$. But then by continuity $f$ must be constant equal to 1, a contradiction.

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    $\begingroup$ It would also work to let $\mu_n$ be a Dirac measure at $1/n$, which would force $f(0)=1$. $\endgroup$ – Nate Eldredge Feb 5 '18 at 15:03

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