There is a question in a syllabus called "Mathematische statistiek":
"If every $X_n$ and $X$ possess a discrete distribution supported on a finite set of integers, show that $X_n$ converges in distribution to $X$ iff $P(X_n=x)\rightarrow P(X=x)$ for every $x$."
This is what I tried: First assume the convergence holds. Let $k_1<k_2<...<k_n$ be all the integers $X$ can reach. For $i\in\mathbb{N}-\lbrace 1\rbrace$ we have \begin{align} P(X_n \leq k_i)-P(X_n\leq k_{i-1})\rightarrow P(X\leq k_i)-P(X\leq k_{i-1})=P(X=k_i) \end{align} But $X_n$ may have other integers in its codomain than $X$ has. So it is not true in general that \begin{align} P(X_n\leq k_i)-P(X_n\leq k_{i-1})=P(X_n=k_i) \end{align} What's more, there is a problem in continuity so that convergence maybe does not hold. The syllabus says "A sequence of random vectors $X_n$ is said to converge in distribution to a random vector $X$ if $P(X_n\leq x)\rightarrow P(X\leq x)$ holds for every $x$ at which the distribution function $x\rightarrow P(X\leq x)$ is continuous."
I hope someone can help. This is the beginning of the subject "Asympotic statistics", so I have only seen the definition of weak convergence and no theorems yet.
$$\begin{multline}P(X_n=k_i) =P(X_n\leq k_i+0.1)-P(X_n\leq k_i-0.1)\to \\ \to P(X\leq k_i+0.1)-P(X\leq k_i-0.1)=P(X=k_i)\end{multline}$$ since the unique integer value $k_i$ belongs to the interval $(k_i-0.1, k_i+0.1]$. The convergence takes place since CDF of $X$ is continuous at the points $k_i\pm 0.1$.