I'm studying probability theory using Kai Lai Chung's textbook.
I was able to understand convergence in probability and almost sure convergence. However, I have hard time understanding the following on vague convergence.
If a sequence of r.v.'s $\{X_n\}$ tends to a limit, the corresponding sequence of p.m's $\{u_n\}$ ought to tend to a limit in some sense. It is true that $\lim\limits_{n} u_n(A)$ exists for all $A \in \mathcal{B}^1$ or at least for all intervals $A$? The answer is no from trivial examples.
Example 1:
Let $X_n = c_n$ where the $c_n$'s are constants tending to zero. Then $X_n \rightarrow 0$ deterministically. For any interval $I$ such that $0 \notin \overline{I}$, where $\overline{I}$ is the closure of $I$, we have $\lim\limits_{n} u_n(I) = 0 = u_n(I) $ ; for any interval such that $0 \notin {I^\circ}$, where $I^\circ$ is the interior of $I$, we have $\lim\limits_{n} u_n(I) = 1 = u_n(I) $. But if $\{c_n\}$ oscillates between strictly positive and strictly negative values, and $I = (a, 0)$ or $(0,b)$, where $a < 0 < b$, then $u_n(I)$ oscillates between $0$ and $1$, while $u(I) = 0$. On the other hand, if $I = (a, 0]$ or $[0,b)$, then $u_n(I)$ oscillates as before but $u(I) = 1$. Observe that $\{0\}$ is the sole atom of $u$ and it is the root of the trouble.
Instead of the point masses concentrated at $c_n$, we may consider, e.g., r.v.'s $\{X_n\}$ having uniform distribution over intervals $(c_n,c^{'}_n)$ where $c_n < 0 < c^{'}_n$ and $c_n \rightarrow 0$, $c^{'}_n \rightarrow 0$. Then again $X_n \rightarrow 0$ a.e. but $u_n((a,0))$ may not converge at all, or converge to any number between $0$ and $1$.
Next, even if $\{u_n\}$ does converge in some weaker sense, is the limit necessarily a p.m.? The answer is again no.
Questions:
1) How do I intepret this? $\lim\limits_{n} u_n(I) = 0 $. Does it mean $u_1(I), u_2(I), u_3(I),...$ to infinity and converge to $0$? Does this interval $I$ stays the same or changes as the sequence $u_n$ goes to infinity?
2) Could someone explain this to me? If $\{c_n\}$ oscillates between strictly positive and strictly negative values, and $I = (a, 0)$ or $(0,b)$, where $a < 0 < b$, then $u_n(I)$ oscillates between $0$ and $1$, while $u(I) = 0$. I have difficulty understanding how $\{c_n\}$ affects $u_n(I)$ oscillates between $0$ and $1$.
3) $0 \notin \overline{I}$ and $0 \notin {I^\circ}$ confused me. Isn't $x \notin \overline I \implies x \notin I^{\circ} $? If $x \notin \overline I$, $\lim\limits_{n} u_n(I) = 0 = u_n(I) $. But
$x \notin \overline I \implies x \notin I^{\circ} $ which then $\lim\limits_{n} u_n(I) = 1 = u_n(I) $ isn't that a contradiction?
Yes, $I$ stays the same as $n\to\infty$. And yes, $\lim_{n} u_n(I)=0$ means $u_n(I)$ will eventually fall into an arbitrarily small neighborhood of $0$ that is previously given.
Since $X_n=c_n$ and the probability measure $\mu_n$ induced by $X_n$ is the point mass measure and its distribution function is $\delta_{c_n}(x)=\begin{cases}0,x<c_n\\ 1, x\ge c_n\end{cases}$. Thus, if $I=(a,0), a<0$ and $\displaystyle c_n=(-1)^n\frac{a}{2} $ for all $n\in\mathbb Z_{>0}$, we have $\mu_n(I)=\frac 12(1+(-1)^{n})$. In other words, $\mu_n(I)$ oscillates betweeen $0$ and $1$.
On page 64 of Kai Lai Chung's book, he claims that: for any interval such that $0\in I^\circ$, where $I^\circ$ is the interior of $I$, we have $\lim_n\mu_n(I)=1=\mu(I)$. So there is no contradiction.