For a sequence $\mu_{n}$ of Borel Probability measures, does $\int f\,d\mu_{n}$ converging for all $f\in C_{b}(\Bbb{R})$ imply that $\mu_{n}$ is a tight sequence?
This is pertaining to the question here. Users(with sufficient privileges) can view my deleted answer where I made the horribly stupid mistake of approximating $\mathbf{1}_{[-M,M]}$ under the supremum norm by continuous bounded functions .
The condition reeks of an application of Uniform Boundedness Principle , but it only yields a bound with the $L^{\infty}$ norm of the form $\sup_{n}|\int f\,d\mu_{n}|\leq C||f||_{L^{\infty}}$ which is not very helpful as we would like to approximate $\mathbf{1}_{[-M,M]}$ by a sequence say $g_{k}$ of continuous bounded functions . So ideally, a bound with the $L^{1}$ norm is what would do the job. Then we can approximate as $k\to\infty$ (uniformly in $n$) $\mu_{n}[-M,M]$ with $\int g_{k}\,d\mu_{n}$ and then use Cauchyness of the sequence $\int g_{k}\,d\mu_{n}$ to prove that $\mu([-M,M])$ is tight.
However, I find no easy way of countering this . I might be having a brain freeze so please excuse my stupidity if I am missing something very easy.
Coincidentally, I've been struggled with a similar problem (which was in an assignment).
・Assuming the contrary, $\exists\epsilon>0$ s.t. $\forall K\subset\mathbb{R}$:compact, $\exists n\in\mathbb{N},\mu_n(K)<1-\epsilon$. In this situation, Show that $\{n\in\mathbb{N};\mu_n(K)<1-\epsilon\}$ is an infinite set.
・Construct $\{\mu_{n_{\ell}}\}$;a subsequence of $\{\mu_n\}$, and $\{K_{\ell}\}$;a sequence of compact subsets of $\mathbb{R}$, s.t. $$ \mu_{n_{\ell}}(K_{\ell})\geq\epsilon \\ |x-y|\geq 1(\forall x\in K_{\ell}, y\in K_m(\ell\neq m)) $$ (Hint:For example, $n_1=1$ and $K_1$ can be taken as $\mu_1(K_1)\geq\epsilon$. $n_2$ can be taken as $\mu_{n_2}(K_1(1))<1-\epsilon$, where $K(1)=\{x\in\mathbb{R};\displaystyle\min_{y\in K}\{|x-y|\}<1\}$. How to choose $K_2\subset\mathbb{R}\setminus K_1(1)$? $n_{\ell}$ and $K_{\ell}$ can be taken inductively.)
・Apploximate the characteristic function of $K_{\ell}$ by continuous bounded functions. (I write that as $f_{\ell}$.) It is expected that $$\int f_{\ell}d\mu_{n_{\ell}}\geq\mu_{n_{\ell}}(K_{\ell})\geq\epsilon.$$ Obviously if there exists $\displaystyle\lim_{n\to\infty}\int f d\mu_n$, for any $\{n_m\}$(:subsequence of $\mathbb{N}$, $n_m\to\infty(m\to\infty)$), $\displaystyle\lim_{n\to\infty}\int f d\mu_n=\lim_{m\to\infty}\int f d\mu_{n_m}$. Derive a contrudiction using $\displaystyle\lim_{n\to\infty}\int 1 d\mu_n=1$, and $\forall N,\exists g_1,\ldots,g_N\in C_b(\mathbb{R})$ s.t. $g_1+\ldots+g_N\leq 1,\displaystyle\int g_i d\mu_{n^i_{m}}\geq\epsilon(\forall m)$.
Sorry for my terrible English. Also, this is my first answer at stackexchange, so my post might be against the rules of this website. For more, please show here (pp.362-366, I reffered to this for my assignment).