Let $X_n \sim U([-1/n,1/n])$ be uniform random variables on $[-1/n,1/n]$ for $n \in \mathbb{N}$. Do the $X_n$ converge, and if yes in what sense?
I think it converges pointwise as for any $x \in [-1/n,1/n]$ you can choose $n > x$ so that $X_n(x)$ is then equal to $0$. Hence $X_n \rightarrow 0$ pointwise.
I believe if you take the measure to be the Lebesgue measure. Then $X_n \rightarrow 0$ in measure as $\lambda(X_n)=2/n \rightarrow 0$.
Does it converge almost everywhere? Not sure how to attempt to write up a proof of this.
What we know is the distribution of $X_n$, but not necessarily $X_n(\omega)$ for each $\omega$. For example, if $Y_n=X_n$ on a set of probability $1$, then $Y_n$ is distributed as $X_n$. In particular, we cannot deduce a pointwise result.
Since for almost every $\omega$ and each $n$, $|X_n(\omega)|\leqslant 1/n$, we obtain that $X_n\to 0$ almost everywhere.