$\sum X_i < \infty$ is a tail event, $X_i$'s are independent, so by the Kolmogorov $0-1$ theorem, $P(\sum X_i < \infty)$ can be either $0$ or $1$. But which one is it?
I tend to think it's $0$.
My logic is as follows:
$X_i\sim U(0,1)$
Let's find the $P(\lim \sup {X_i>0.5})$. Using Borel-Cantelli, since the sum $\sum_{i=1}^\infty P(X_i > 0.5) = \sum_{i=1}^\infty 0.5 = \infty$ and since the series of R.V. are independent, then $P(\lim \sup {X_i>0.5})=1$.
Then $\sum_{i=1}^\infty X_i > \sum_{\lim \sup {X_i>0.5}}X_i>\sum_{\lim \sup {X_i>0.5}}\frac{1}{2}=\infty $
i.e. we have infinite elements in the series that will be larger than $0.5$, and so the series will not converge, hence the probability will be $0$.
Is that correct, or am I missing something?
The argument is fine, but I would write a bit differently the index of summation $\sum_{\lim \sup {X_i>0.5}}X_i$. Instead, one can define $I=I(\omega)=\{i\geqslant 1, X_i(\omega)>0,5\}$ and write $$ \sum_{i=1}^\infty X_i(\omega) \geqslant \sum_{i\in I(\omega)}X_i(\omega) $$ which is almost surely infinite, as $I(\omega)$ is almost surely infinite and $X_i(\omega)>0.5$ for $i\in I(\omega)$.