I was wondering, if you have a sequence of probability measures $(\mu_n)_n$ on $\mathbb R$ and you know that there is a probability measure $\mu$ such that for all $k\in\mathbb N=\{0,1,2,\cdots\}$ $ \lim_{n\rightarrow\infty}\int x^kd\mu_n(x)=\int x^kd\mu(x), $ does it imply that for any continuous and bounded function $f$ you have $ \lim_{n\rightarrow\infty}\int f(x)d\mu_n(x)=\int f(x)d\mu(x) \qquad ? $ And if no, what if $\mu$ has compact support ?
EDIT : I'm kind of lost : I understand reading the answers that it is somehow necessary that $\mu$ is characterized by its moments, but on an other hand, I come up with this proof, where I don't see what's wrong with. Could you help ?
Since $\mu_n$ converges towards $\mu$ in moments, we have by density of the polynomials in $C_c(\mathbb R)$ that for any $h\in C_c(\mathbb R)$ $ \lim_{n\rightarrow\infty}\int h(x)d\mu_n(x)=\int h(x)d\mu(x). $ Now, let $f\in C_b(\mathbb R)$, take any $h\in C_c(\mathbb R)$ satisfying $0\leq h \leq 1$, and write $ \left|\int f(x)d\mu_n(x)-\int f(x)d\mu(x)\right| $ $ \leq \left|\int f(x)d\mu_n(x)-\int f(x)h(x)d\mu_n(x)\right|+\left|\int f(x)h(x)d\mu_n(x)-\int f(x)h(x)d\mu(x)\right|+\left|\int f(x)h(x)d\mu(x)-\int f(x)d\mu(x)\right|. $ Thus, using $\lim_n\mu_n(\mathbb R)=\mu(\mathbb R)$ (i.e the convergence of the "$0$-th moment"), we obtain $ \limsup_n\left|\int f(x)d\mu_n(x)-\int f(x)d\mu(x)\right|\leq 2\|f\|_{\infty}\int(1-h(x))d\mu(x). $ Finally, given $\epsilon >0$, one can choose $g$ such that $ \int(1-h(x))d\mu(x)\leq \frac{\epsilon}{2\|f\|_{\infty}}, $ I have the impression that I obtain the result... Where's the mistake ?
Thanks in advance !