Could we use DCT with pointwise convergence in probability? I'm interested in finding the limit in probability of an integral
$$
\int_{0}^\infty f(X_n, t) \, \pi(\mathrm{d} t),
$$
where $X_n$ is a sequence of random variables and $\int_{0}^\infty \pi(\mathrm{d} t) = 1.$
I know the pointwise limit in probability of $f(X_n, t)$ and I can bound the function $|f(x, t)|$ by a function $h(t)$ that is integrable $\pi( \mathrm{d} t)$. Is that enough to bring the probability limit inside the integral?
More precisely, I have
\begin{align*}
f(X_n, t) &\rightarrow_P f(X, t) \\
|f(x, t)| &\le h(t),\, \, \int_{0}^\infty h(t) \, \pi(\mathrm{d} t) < \infty
\end{align*}
Is it true that
$$
\int_{0}^\infty f(X_n, t) \, \pi(\mathrm{d} t) \rightarrow_P \int_{0}^\infty f(X, t) \pi(\mathrm{d} t) ?
$$
My thoughts: It looks like I should be able to apply the dominated convergence theorem (DCT), somehow. I'm confused because it differs from the textbook examples I saw when I took a basic course in analysis. For example, I'm taking a limit in probability outside the integral. I haven't applied DCT like that before: the limit outside the integral was a "regular" limit.
References or answers are both appreciated. Thanks!
 A: Let $\epsilon > 0$
\begin{align}
\left|\int_0^\infty \left(f\left(X_n, t\right) - f(X,t)\right)\pi(\mathrm dt)\right| &\le \frac\epsilon2 \int_{0}^\infty \mathbf 1_{\left|f\left(X_n, t\right) - f(X,t)\right| \le \frac\epsilon 2}(t)\pi(\mathrm dt) + \int_{0}^\infty \left|f\left(X_n, t\right) - f(X,t)\right|\mathbf 1_{\left|f\left(X_n, t\right) - f(X,t)\right| \ge \frac\epsilon 2}(t)\pi\left(\mathrm dt\right)\\
&\le \frac\epsilon 2 + 2\int_0^\infty h(t)\mathbf 1_{\left|f\left(X_n, t\right) - f(X,t)\right| \ge \frac\epsilon 2}(t)\pi(\mathrm dt)
\end{align}
With that we have:
\begin{align}
\mathbb P\left[\left|\int_0^\infty \left(f\left(X_n, t\right) - f(X,t)\right)\pi(\mathrm dt)\right| \ge \epsilon\right] &\leq \mathbb P\left[\frac\epsilon 2 + 2\int_0^\infty h(t)\mathbf 1_{\left|f\left(X_n, t\right) - f(X,t)\right| \ge \frac\epsilon 2}(t)\pi(\mathrm dt) \ge \epsilon\right]\\
&= \mathbb P\left[2\int_0^\infty h(t)\mathbf 1_{\left|f\left(X_n, t\right) - f(X,t)\right| \ge \frac\epsilon 2}(t)\pi(\mathrm dt) \ge \frac\epsilon 2\right]\\
&= \mathbb P\left[\int_0^\infty h(t)\mathbf 1_{\left|f\left(X_n, t\right) - f(X,t)\right| \ge \frac\epsilon 2}(t)\pi(\mathrm dt) \ge \frac\epsilon 4\right]\\
&\le \frac{\mathbb E\left[\displaystyle\int_0^\infty h(t)\mathbf 1_{\left|f\left(X_n, t\right) - f(X,t)\right| \ge \frac\epsilon 2}(t)\pi(\mathrm dt)\right]}{\frac\epsilon 4}\\
&=\frac{\displaystyle\int_0^\infty h(t)\mathbb E\left[\mathbf 1_{\left|f\left(X_n, t\right) - f(X,t)\right| \ge \frac\epsilon 2}(t)\right]\pi(\mathrm dt)}{\frac\epsilon 4}\\
&= \frac{\displaystyle\int_0^\infty h(t)\mathbb P\left[{\left|f\left(X_n, t\right) - f(X,t)\right| \ge \frac\epsilon 2}\right]\pi(\mathrm dt)}{\frac\epsilon 4}\\
&= \frac4\epsilon \int_0^\infty g_n(t)\pi(\mathrm dt)
\end{align}
where $g_n(t) = h(t)\mathbb P\left[{\left|f\left(X_n, t\right) - f(X,t)\right| \ge \frac\epsilon 2}\right]$. By the assumption, you have $g_n(t)$ converges to $0$ pointwise and since $g_n(t) \le h(t)$, you can use DCT to have that:
$$\int_0^\infty g_n(t) \pi(\mathrm d t) \to 0$$
which proves what you are aiming for.
