If they exist, are the limits of "almost sure convergence" and "convergence in mean" equal for a sequence of random variables?

Question

I have a sequence of random variables $(X_n)_{n \in \mathbb{N}_0}$ which converge both "almost sure" and "in mean" to random variables $A$ and $B$: $$ P(\lim X_n = A) = 1 \text{ (almost sure convergence)} \\ E(|X_n - B|) \rightarrow 0 \text{ (convergence in mean)} $$

My guess would be that $A = B$ almost sure, but I couldn't come up with a proof.

I've already looked on the Wikipedia page about stochastic convergence but I couldn't find any information about this particular case there.

Do you know any results concerning this matter?

Which additional properties does $(X_n)_n$ need to have such that $A = B$ almost sure?

Do you know any examples in which $P(A \neq B) > 0$?

Answer 1

Both types of convergence imply convergence in probability. Hence $X_n$ converges in probability to both $A$ and $B$. And it is well known that two limits in probability are equal almost surely. Hence $P(A=B)=1$.

Answer 2

If $X_n \to B$ in mean, then there is a subsequence $X_{n_k}$ that converges a.s. to $B$. If also $X_n \to A$ a.s. then of course that subsequence converges a.s. to $A$. If a sequence $X_{n_k}$ converges a.s. to both $A$ and $B$, then $A=B$ a.s.