Showing that a sequence of random variables with $\mathcal{L}(X_{n})$ (I.e., law) uniform on $[-n,n]$ does not converge at all Let $(X_{n})_{n\geq 1}$ be a sequence of real valued random variables with $\mathcal{L}(X_{n})$ (that is, law or distribution) uniform on $[-n,n]$. In what sense(s) do $X_{n}$ converge to a random variable $X$?
We are told that the answer is none. To prove this, are we supposed to show that $X_{n}$ does not converge under the weakest form of convergence possible (convergence in distribution/law)? And if we do, is that sufficient to show that there is no convergence? Also, how does the fact that the r.v.'s are uniformly distributed give us that?
 A: Convergence in distribution means
$$\lim_{n\rightarrow \infty}F_{X_n}(x) = F_X(x)$$
where the RHS is a distribution function, and the equality to hold for every $x$ for which  $F_X(x)$ is continuous.
Our distribution function is  
$$F_{X_n}(x)= \begin{cases}0&\text{if $x<-n$}\\\frac{x+n}{2n}&\text{if $x\in[-n,n]$} \\1&\text{if $x> n$.}\end{cases}$$
The first and the last branch are not defined as $n\rightarrow \infty$, since there is no $x$ lower than "minus infinity", or higher than "plus infinity". Then
$$\lim_{n\rightarrow \infty}F_{X_n}(x) = \lim_{n\rightarrow \infty}\Big (\frac{x}{2n} + \frac 12\Big) = 0+\frac 12$$
The constant function $1/2$ does not satisfy the properties of a distribution function, specifically
$$\lim_{x\rightarrow -\infty}\frac 12 =\frac 12 \neq 0,\;\;\lim_{x\rightarrow \infty}\frac 12 =\frac 12 \neq 1$$
Carefully note that $n$ and $x$ do not "go together" at plus/minus infinity. To study convergence we first "send $n$ to infinity", and then we examine the behavior of the limiting function we have obtained, as its argument passes over.
A: If $U$ is uniform on $[-1,1]$, then $nU$ is uniform on $[-n,n]$. We have for each $t$,
$$\mathbb P(nU\leqslant t)=\mathbb P(U\leqslant t/n)\to 1/2,$$
since the cumulative distribution function of $U$ is continuous at $0$. 
This proves that $(nU)_n$ cannot converge in distribution. 
