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Let $X_n \sim U(0, 1/n)$ and $X = \delta_0$, that is, $P(X = 0) = 1$. I found arguments for $X_n \stackrel{d}{\to} X$ and $X_n \not\stackrel{d}{\to} X$ which contradict each other.

Let $F_n$, $F$ be the corresponding distribution functions. We now have $$ F_n(t) = \begin{cases} 0, & \text{if $t \leq 0$} \\ nt, & \text{if $0 < t < 1/n$} \\ 1, & \text{if $1/n \leq t$} \end{cases} $$ Thus we have $F_n(t) \to F(t)$ pointwise (because $1/n < t$ for large $n$) except for $t=0$, which is the only value where $F$ isn't continuous. Thus we may write $X_n \stackrel{d}{\to} X$.

On the other hand, $\cos$ is continuous and bounded. We have $$ E(\cos(X_n)) = \int_0^{1/n} \cos(x) dx = \sin(1/n) \to 0 $$ but also $$ E(\cos(X)) = P(X=0)\cdot \cos(0) = 1. $$ It follows that $X_n \not\stackrel{d}{\to} X$. Where am I going wrong?

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    I think that $$ F(t) = \begin{cases} 0, & \text{if $t \leq 0$} \\ nt, & \text{if $0 < t < 1/n$} \\ 1, & \text{if $1/n \leq t$} \end{cases} $$ should be $F_n(x)$ instead of $F(x)$. Am I correct? – 2017-02-22
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    @flytothesurface You are indeed correct, thank you very much. – 2017-02-22

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Your first statement, which is exactly the definition of convergence in distribution is correct.

You counter based on $E(\cos(X_n)$ is flawed because the probability density function $f_n(t) = n$ on $(0,1/n)$ and you used $f_n(t) = 1$ in that interval. Your correct calculation would read $$ E[\cos(X_n)] = \int_0^{1/n}n\cos(x)\,dx = n\sin(\frac1n) \\ \lim_{n\to\infty} n\sin(\frac1n) = 1 \\ \lim_{n\to\infty} E[\cos(X_n)] = 1 = E[\cos(X)] $$ so this probe does not rule out convergence in distribution.

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    ... which turns this into a cool way to prove $\lim_{n\to\infty} n\sin(\frac1n) = 1$. Awesome! – 2017-02-22
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    Ah, on second thought I think that limit is usually used to show (sin)' = cos, which one has to use. Oh well. Thank you though! – 2017-02-22