Q:

Convergence of random variables taking integer values

Let $X_n$ be random variables taking integer values, and let $X_n\to X$ in distribution.

Show $X$ also takes only integer values.
$P(X_n=j)\to P(X=j)$ for each integer $j$.
$\displaystyle \sum_{j}|P(X_n=j)-P(X=j)| \to 0$ as $n\to \infty$

My try :
WLOG assume $j$ varies over $1,2,\dots$ Now take $x$ which is not an integer. Consider an interval $(x_1,x_2]$ containing $x$, but no integers, and $x_1,x_2$ continuity points of $F_X$. Now $P(x_1<X_n\le x_2)=0\to P(x_1<X\le x_2)$. Now if $X_n$ took non-integral values the probability on rhs would be nonzero, but that is a contradiction. For the second we do something similar. Take $x=j$ and get $j_1,j_2$ as previous. Now taking $j_1\to j^-$ and $j_2\to j^+$ we have our desired conclusion. I think the first two proofs are correct but what about the third one? Can someone help me? Thanks.

A:

(1) follows from 'Portmanteau Theorem'. Denote probability measure induced by $X_n,X$ by $F_n$ and $F$ respectively. Note $\mathbb{Z}$ is closed in $\mathbb{R}$. As $X_n \overset{d}{\rightarrow} X$ we've $F(\mathbb{Z}) \geq \limsup F_n(\mathbb{Z})=1$ as $F_n(\mathbb{Z})=1$ $\forall n$
(2) You can say something strong namely $P(X_n=j) \rightarrow P(X)$ as $n\rightarrow \infty$ uniformly in the intergs $j$  i.e. $\sup_{\mathbb{Z}} |P(X_n=j) - P(X)|\rightarrow 0$ as $n\rightarrow 0$. How?
First show that for any integer valued random variable Y whith characteristic function $\phi$
$$ P(Y=j)=\frac{1}{2\pi} \int_{-\pi}^{\pi} e^{-ijt}\phi(t) dt$$
Then since $X_n \overset{d}{\rightarrow} X$ we also have $\phi_n(t) \overset{u}{\rightarrow} \phi(t) \forall t$. Also note $|e^{-ijt}|=1$ where $i=\sqrt{-1}$ and
$$|P(X_n=j) - P(X=j)|\leq \frac{1}{2\pi} \int_{-\pi}^{\pi} |\phi_n(t)-\phi(t)| dt \hspace{10pt} \text{RHS does not depend on $j$}$$
Use Lebesgue DCT.
(3) You can even say more! This is (I think) known as Scheffe's Theorem for p.m.f's. We've from part (2) 
$p_n(j) \rightarrow p(j)$ $\forall j\in \mathbb{Z} $ 
(Here I'm denoting $P(X_n=j)=p_n(j)$ and $P(X=j)=p(j)$).
Need to show $\sum |p_n(j)-p(j)|\rightarrow 0$ as $n\rightarrow \infty$.Now
$$\limsup_{n\rightarrow \infty} \sum_{j=1}^{\infty}|p_n(j)-p(j)|\leq \limsup \Bigl(\sum_{j=1}^N |p_n(j)-p(j)|\Bigr)+\limsup \sum_{j=N+1}^{\infty} |p_n(j)-p(j)|=\limsup \Bigl(\sum_{j=N+1}^{\infty} |p_n(j)-p(j)|\Bigr) \hspace{10pt} \forall N\geq 1$$
Thus suffices to show $\limsup \Bigl(\sum_{j=N+1}^{\infty} |p_n(j)-p(j)|\Bigr)< \epsilon$ for any $\epsilon >0$.
As $\sum p(j)=1$ we can have $\sum_{j=N+1}^{\infty} p(j)<\epsilon/2$ for large enough $N$.
For that $N$, 
$$\hspace{10pt}\sum_{j=N+1}^{\infty} p_n(j)=1-\sum_{j=1}^N p_n(j)\rightarrow 1-\sum_{j=1}^N p(j)=\sum_{j=N+1}^{\infty} p(j)<\epsilon/2$$
Thus we get, $\limsup \sum|p_n(j)-p(j)|<\epsilon/2 +\epsilon/2=\epsilon$

