Suppose that $X_n \stackrel{P}{\rightarrow} c$ as $n\to\infty,$ where $X_n, c$ are all positive. Let $f$ be a continuous (but possibly unbounded) function on $(0,\infty)$ such that $$\int_0^\infty |f(x)|^pdx <\infty$$ for some $p>1.$ Then is it true that $$\mathbb{E}\left(\left|f(X_n)- f(c)\right| \mathbb{1}_{|X_n - c|>\delta}\right)\to 0\text{ as }n\to\infty?$$ The above is immediate for bounded $f$, but note that here we are not imposing that condition.
In case the above conclusion does not hold in general, another question would be: does it hold under the assumption that $\text{Var}(X_n)\to0$ as $n\to\infty$?
This issue arose while generalising a proof of Stone-Weierstrass theorem (the usual statement is given for bounded continuous $f$ only). Any help is appreciated.
I doubt whether $f(X_{n})$ is integrable.
Consider the following example: Let $(\Omega,\mathcal{F},P)$ be a probability space, where $\Omega=(0,1)$, $\mathcal{F}=\mathcal{B}((0,1))$, and $P$ is absolutely continuous with repsect to the Lebesgue measure $m$ and defined by $dP(\omega)=\frac{1}{2}\frac{1}{\sqrt{\omega}}dm(\omega).$ Let $f:(0,\infty)\rightarrow\mathbb{R}$ be defined by $f(x)=\frac{1}{\sqrt{x}}1_{(0,1)}(x)+\frac{1}{x^{2}}1_{[1,\infty)}(x)$ and $p\in(1,2)$, then $f$ is continuous and $\int_{0}^{\infty}|f(x)|^{p}dx<\infty$. Let $c=\frac{1}{2}.$ For each $n$, define $X_{n}:\Omega\rightarrow(0,\infty)$ by $X_{n}(\omega)=\omega1_{(0,\frac{1}{n})}(\omega)+c1_{[\frac{1}{n},1)}(\omega).$ For each $\delta>0$, $\{\omega\in\Omega\mid|X_{n}(\omega)-c|\geq\delta\}\subseteq(0,\frac{1}{n})$, so $P\{\omega\in\Omega\mid|X_{n}(\omega)-c|\geq\delta\}\leq P\left((0,\frac{1}{n})\right)=\frac{1}{\sqrt{n}}\rightarrow0$. This shows that $X_{n}\rightarrow c$ in probability.
Next, we argue that $f(X_{n})$ is not integrable. Observe that \begin{eqnarray*} \int f(X_{n})dP & = & \int_{0}^{1}f(X_{n}(\omega))\frac{1}{2\sqrt{\omega}}dm(\omega)\\ & \geq & \int_{0}^{\frac{1}{n}}f(\omega)\cdot\frac{1}{2\sqrt{\omega}}dm(\omega)\\ & = & \int_{0}^{\frac{1}{n}}\frac{1}{2\omega}dm(\omega)\\ & = & \infty. \end{eqnarray*}