If $X_n$ is a sequence and uniformly distributed on $(-n,n)$. Does it converge in $L^p$ and if it does then what would be the limit?
MY attempt consisted of finding the characteristic function which leads us to
the characteristic function $$ \varphi_{X_n}(t) = \frac{\sin(tn)}{tn}$$ and so:
$$\varphi_{X_n}(t)=\begin{cases}\frac{\sin(tn)}{tn} &t \ne0\\ 1&t = 0\end{cases}.$$
and then: $$\lim_{n\to\infty} \varphi(t)=\begin{cases}0 &t \ne0\\ 1&t = 0\end{cases}.$$
Now, I can see that when $t = 0$ for $t \in \mathbb{R} $, the limiting characteristic function $\varphi(t)$ is discontinuous. So the limit exists except for when $t=0$. Hence we see that there is no convergence in distribution. From here can we conclude that it does not converge in $L^p$?
Is there any other way to prove this. Thanks
$$E|X_n|^{p}=\frac 1 {2n}\int_{-n}^{n} |t|^{p}dt$$ $$ \geq \frac 1 {2n}\int_{n/2}^{n} |t|^{p}dt$$ $$ \geq \frac 1 {2n} {\frac n 2} |\frac n 2|^{p} \to \infty$$ as $n \to \infty.$ Hence $(X_n)$ is not even bounded in $L^{p}$ for any $p >0$.