Prove or disprove: convergence in distribution of continuous uniformly distributed variables Let $(X_n)_{n\in\mathbb{N}}$ be a sequence of random variables, with $X_n\sim U[-n,n]$ for all $n\in\mathbb{N}$.
Prove or disprove: this sequence converges in distribution.
I am not sure if this is true. My first attempt is to work with characteristic function. For $n\in\mathbb{N}$,
$$\varphi_{X_n}(t)=\frac{\sin(nt)}{nt}$$ which approaches $0$ for $t\neq0$ as $n\rightarrow\infty$ and $1$ otherwise. I think I need to use
Lévy's continuity theorem: If a sequence of characteristic functions converges pointwise to a limiting function $\varphi$ which is continuous at zero then $\varphi$ is a characteristic function and the sequence of random variables converges in distribution.
Problem is: What I am not quite sure about is whether or not the limiting function discontinuous at $0$?
Thanks in advance!
 A: There are a few ways to do this. One way is to work with the ChF (characteristic function) and appeal to the theorem you quoted. For $t\ne 0$, as $n\to\infty$ we have
$$|\varphi_{X_n}(t)|=\left|\frac{\sin(nt)}{nt}\right|\le \frac{1}{nt}\to 0$$
while for $t=0$, recall that the ChF of any random variables at $t=0$ is $1$. Thus, $\varphi_{X_n}(0)=1$ by definition for all $n\in\mathbb{N}$ and so $\varphi_{X_n}(0)\to 1$ as $n\to\infty$. This means that $\varphi_{X_n}(t)\to \boldsymbol{1}_{\{0\}}(t)$, namely the function which is $1$ at $t=0$ and zero everywhere else. This function is discontinuous at $t=0$ since $\boldsymbol{1}_{\{0\}}(t)\to0 \ne \boldsymbol{1}_{\{0\}}(0)$ as $t \to 0$. If you were to draw the graph out, you'll see that there is a sudden jump at $t=0$. In any case, the sequence does not converge in distribution.
The second way to do this is to work directly with the DF (distribution function). Let $F_n$ be the DF of $X_n$ and suppose otherwise that $X_n\overset{d}{\to}X$ for some $X$ with DF $F$ as $n\to\infty$. See that $F_n(x)\to 1/2$ for all $x\in\mathbb{R}$ since $F_n(x)=1$ for $x>n$, $F_n(x)=0$ for $x<-n$ and for $x\in [-n,n]$, $$F_n(x)=\frac{1}{2n}x+\frac{1}{2}$$ Therefore, $F(x)=1/2$ for all $x\in\mathbb{R}$ (this may not be true for at most countably many points, but since $F$ is right-continuous, this is true everywhere). But the constant function $1/2$ is not a DF and so $F$ is not a DF, contradicting the fact that $F$ is the DF of $X$. Therefore, $(X_n)_{n=1}^\infty$ does not converge in distribution.
