Let $X_n$ $(n=1,2,\dots)$ be a sequence of discrete random variables, where the distribution of $X_n$ is the discrete uniform over $\{0, 1/n, 2/n,\dots,1 \}$. Let $U$ be a random variable whose distribution is the continuous uniform over $[0,1]$.
Does $X_n \to U$ almost surely?
My approach is to use Borel-Cantelli to check that $\sum_n\Pr[|X_n-U|>\epsilon] < \infty$, but it appears that for $n>2$ and $\epsilon_n < 1/(n+1)$, we have $\Pr[|X_n-U|>\epsilon_n] = 1-2n{\epsilon}_n/(n+1)$, whose sum does not converge. Is there another approach?
If the convergence does not happen almost surely, is it possible for any discrete random variable to converge almost surely to a continuous random variable? (It is clear in the example above that that $X_n \to^\mathrm{D} U$ i.e. convegence in distribution.)
As written, the statement is false. Suppose the sample space is $\Omega = [0,1]$, with the probability measure $P$ being the Lebesgue measure. Define the random variables $$ X_n : \Omega \to \mathbb R, \ \ \ \ X_n(\omega)=\begin{cases} \frac{\lfloor (n+1)\omega\rfloor}{n} & {\rm if \ } n {\rm \ is \ odd} \\ 1 - \frac{\lfloor (n+1)\omega\rfloor}{n} & {\rm if \ } n {\rm \ is \ even} \end{cases}$$
Notice that the distribution of $X_n$ is discrete uniform over $\{0, 1/n, 2/n, \dots, 1 \}$, and this is true regardless of whether $n$ is odd or even.
And yet, $X_n$ fails to converge pointwise a.s.