Let $M_n$ be a sequence of random variables and let $x\in\mathbb{R}$ such that $$\mathbb{E}[h(M_n)]\to h(x)$$ as $n\to\infty$ for all bounded and continuous functions $h$.
How to show that for all $\delta>0$ we have $\mathbb{P}(|M_n-x|>\delta)\to0$?
What I thought:
Convergence of $\mathbb{E}[h(M_n)]$ means that for all $\epsilon>0$ there is an $N\geq0$ such that $\mathbb{E}[h(M_n)-h(x)]<\epsilon$.
Further we hav $\mathbb{P}(|M_n-x|>\delta)=\mathbb{E}[1_{|M_n-x|>\delta}]$, however the indicator function is not continuous. How do I fix this?
The stated assumption implies that $M_n\xrightarrow{d}x$ and, thus, $M_n\xrightarrow{p} x$.
Direct proof. For any $\delta>0$, \begin{align} \mathsf{P}(|M_n-x|>\delta)&=\mathsf{P}(M_n>x+\delta)+\mathsf{P}(M_n<x-\delta) \\ &\le 1+\mathsf{E}h_{\delta}(M_n-x)-\mathsf{E}h_{\delta}(M_n+\delta-x),\tag{1} \end{align} where $h_\delta(v):=1\wedge(0\vee v/\delta)$. Since $h_{\delta}$ is a continuous bounded function, $(1)$ converges to $$ 1+h_{\delta}(0)-h_{\delta}(\delta)=0. $$