Can a sequence of (probability) measures converge to zero? Considering the r.v. sequence $X_n$ ~ $U(-n,n)$, show $X_n$ doesn't converge to a r.v $X$.
I am thinking that $\mu_n(x)=\frac{1}{2n}, x \in (-n,n)$, so $\lim\limits_{n\to ∞}\mu_n(x)=0$. A measure itself cannot be defined $\mu(x):= 0$, correct? Therefore it does not weakly converge, and so does not converge in any other sense.
Is this basically correct? Am I way off by my conceptualization of $\mu$?
 A: The sequence of distribution functions $F_n(x)=\left(\frac{(x+n)}{2n}\wedge 1\right) \vee 0$ converges to $F\equiv 1/2$ for all $x\in\mathbb{R}$, which is not a proper cdf. That is, the sequence $\{X_n\}$ does not converge in distribution (and in other mode stronger than that). In fact, we say that $F_n\to F$ vaguely.
A: Another approach would be to use characteristic functions, and make use of Levy's continuity theorem.
Let $\phi_n(t) := \mathbb{E}(e^{itX_n})$ and $\phi(t)=\mathbb{E}(e^{itX})$ be the characteristic function of $X_n$ and $X$, respectively.
By Levy,  $X_n \stackrel{d} {\rightarrow} X$ if and only if $\phi_n(t) \rightarrow \phi(t)$. Now, suppose $X_n$ converges weakly, i.e., $X_n \stackrel{d} {\rightarrow} X$ for some $X$.
We know that
$$
\phi_n(t) = \frac{e^{itn}-e^{it(-n)}}{2itn},
$$
and if we take the limit $n\rightarrow \infty$,
$$
\begin{split}
\lim_{n\rightarrow \infty} \phi_n(t) &= \lim_{n\rightarrow \infty} \frac{e^{itn}-e^{it(-n)}}{2itn} \\
&=0,
\end{split}
$$
$\phi(t) = 0$ corresponds to the measure that's $0$ a.e.
