It'd be of great help if someone could double-check if my proof is correct/rigorous.
Problem: For any $p>0$, if $X_n\to X$ and $X_n\to Y$ in $L^p$, then $X=Y$ almost surely
My proof: We start with the case of $p\geq 1$. We have that $\lVert X_n -X \rVert_p\to 0$, therefore by Markov's inequality:
$$\mathbb{P}(\lvert X_n -X\rvert>\epsilon)\leq \frac{\mathbb{E}\lvert X_n-X\rvert}{\epsilon}\to 0$$ as $n$ goes to $\infty$. Same thing with $Y$. So then we have $$\mathbb{P}(\lvert X-Y\rvert >\epsilon)\leq \frac{\mathbb{E}(\lvert X-Y\rvert)}{\epsilon}= \frac{\mathbb{E}(\lvert X-X_n + X_n -Y\rvert)}{\epsilon}\leq \frac{\mathbb{E}(\lvert X-X_n\rvert + \lvert X_n -Y\rvert)}{\epsilon}=\frac{\mathbb{E}(\lvert X-X_n\rvert) + \mathbb{E}(\lvert X_n -Y\rvert)}{\epsilon}$$ which converges to $0$ as $n$ goes to infinity. Therefore, we have that $\mathbb{P}(X\neq Y)=0$ and we are done.
As for $0<p<1$, a similar (or rather generalized) argument can be made through showing that $L_p$ convergence yields convergence in probability.
Best Answer
Here are a few things you might want to think about.
First, are you comfortable with the leap from $\|X_n-X\|_p\to 0$ for some specific $p$ to concluding $\|X_n-X\|_1 \equiv\mathbb{E}|X_n-X|\to 0$? If you are trying to use the fact that $L^p$ convergence for any $p\geq 1$ implies $L^1$ convergence, then it would make more sense to simply say "because $X_n$ converges to $X$ in $L_p$ (for some $p\geq 1$), we must have $\mathbb{E}|X_n-X|\to0$." If you are trying to use the actual convergence of the $p$-norm $\|X_n-X\|_p\to0$ to conclude $\mathbb{E}|X_n-X|\to0$, then you should probably invoke Jensen's (or Hölder's) inequality as an intermediate step. Alternatively, you can apply Markov's inequality to the random variable $|X_n-X|^p$ instead of $|X_n-X|$ (and maybe rewrite $\epsilon$ as $\epsilon^p$).
Second, after showing that $P(|X_n-X|>\epsilon)\to 0$ and observing that a similar fact holds for $Y$, you don't actually appear to use these results anywhere. You just re-apply Markov's inequality again directly to $P(|X-Y|>\epsilon)$ and work from there.
Third, in the last steps of your proof, you go from showing that $P(|X-Y|>\epsilon) \leq c_n$ for some $c_n\to 0$ straight to concluding that $P(X\not= Y)=0$. Are you comfortable that the former implies the latter? If I pointed out that $Z_n \to Z$ in $L^p$ implies the following by Markov's inequality: $$P(|Z_n-Z|>\epsilon) \leq \frac{E(|Z_n-Z|)}\epsilon \to 0$$ and therefore we have that $P(Z_n\not= Z)=0$ for sufficiently large $n$, proving that $L^p$ convergence implies almost sure convergence, could you explain why your argument works but mine doesn't?