Convergence of a stochastic process Suppose that $\{X_n(t),t\in\mathbb{R}\}_{n=0,1,...}$ is a collection of stochastic processes, i.e., for any fixed $n$, we have and stochastic process $\{X_n(t),t\in\mathbb{R}\}$. 
Assume that for any fixed $n$ we have that $X_n(t)\overset{d}{\to}N(0,t^2)$ for all $t\in \mathbb{R}$ when $n\to\infty$ ($N(0,0)$ is the constant $0$.) Now, if $(T_n)_{n\geq 0}\subset\mathbb{R}$ is a sequence of random variables  such that $T_n\overset{P}{\to}0$ when $n\to\infty$, is it true that 
$$
X_n(T_n)\overset{P}{\to}0
$$
when $n\to\infty$? This seems true since, intuitively, $X_n(T_n)\sim N(0, T_n^2)$ for large $n$. My only "formal" approach is to calculate 
$$
P(|X_n(T_n)|>\epsilon)=\int_{\mathbb{R}}P(|X_n(t)|>\epsilon|T_n=t)dF_n(t),
$$
where $F_n(t)=P(T_n\leq t)$. Since $F_n(t)\to \mathbf{1}_{[0,\infty)}(t)$ when $n\to\infty$ then $dF_n(t)/dt=0$ for $t\neq 0$. I don't know where to go from here.
Any ideas on how to prove it? If it is not true, is there any other assumption under it is true? 
 A: Assuming that a.s. the family $(X_n)_n$ is equicontinuous at $0$. The convergence might hold under some weaker assumption but I found this quite natural.
Let $\epsilon >0$, then there exists $\delta >0$ such that $0\leq t \leq \delta \Longrightarrow |X_n(t)-X_n(0)| \leq \epsilon$ for every $n$. Now write
$$\mathbb{E}\left|e^{iX_n(T_n)} - 1\right| \leq \mathbb{E}\left|e^{iX_n(0)} - 1\right| + \mathbb{E}\left|e^{iX_n(T_n)} - e^{iX_n(0)}\right|.$$
The first term goes to $0$ since $X_n(0) \to 0$ in distribution. For the second term, we have the decomposition
$$\begin{align}\mathbb{E}\left|e^{iX_n(T_n)} - e^{iX_n(0)}\right|&= \mathbb{E}\left[\left|e^{iX_n(T_n)} - e^{iX_n(0)}\right|\mathbf{1}_{T_n \leq \delta}\right] + \mathbb{E}\left[\left|e^{iX_n(T_n)} - e^{iX_n(0)}\right|\mathbf{1}_{T_n >\delta}\right]\\
&\leq \mathbb{E}\left[\left|e^{iX_n(T_n)} - e^{iX_n(0)}\right|\mathbf{1}_{|X_n(T_n)-X_n(0)|\leq \epsilon}\right] + 2 \mathbb{P}(T_n >\delta) \\
&\leq \mathbb{E}\left[\left|X_n(T_n) - X_n(0)\right|\mathbf{1}_{|X_n(T_n)-X_n(0)|\leq \epsilon}\right] + 2 \mathbb{P}(T_n >\delta)\\
&\leq \epsilon + 2 \mathbb{P}(T_n >\delta).
\end{align}$$
Since $T_n \to 0$ in probability, it follows that
$$\limsup_n \mathbb{E}\left|e^{iX_n(T_n)} - e^{iX_n(0)}\right| \leq \epsilon.$$
This holds for every $\epsilon >0$, so letting $\epsilon \to 0$ gives the convergence in distribution $X_n(T_n) \to 0$.
