What are the conditions on the sequence on $\{X_n\}$ (apart from the degenerate random variable), under which it can be claim that $||X_n-X||_{L^2(\mathbb{R})}\rightarrow 0$ implies $X_n\rightarrow X$, almost surely?
I know that there always exists subsequences along with the above implication hold (as in the second answer of this question). But I want to know about the convergence of the whole sequence by imposing some condition on it.
Thank in advance.
Here are some conditions: Suppose $X$ is a random variable and $\{X_n\}_{n=1}^{\infty}$ are a sequence of random variables.
Claim 1: If for all $\epsilon>0$ we have $\sum_{n=1}^{\infty} Pr[|X_n-X|>\epsilon]<\infty$, then $X_n\rightarrow X$ with probability 1.
Claim 2: Suppose there is a constant $d>0$ such that: $$ \sum_{n=1}^{\infty} E[|X_n-X|^d] < \infty $$ Then the conditions of Claim 1 hold, and so $X_n\rightarrow X$ with probability 1.
Example: Suppose $E[(X_n-X)^2] \leq \frac{5}{n^{1.1}}$ for all $n \in \{1, 2, 3, \ldots\}$. Then $X_n\rightarrow X$ with probability 1, since $\sum_{n=1}^{\infty} \frac{5}{n^{1.1}} < \infty$.
Proof (Claim 1): Fix $\epsilon>0$. We want to show that $\lim_{M\rightarrow\infty} Pr\left[\cup_{n\geq M} \{|X_n-X|>\epsilon\} \right]=0$. By the union bound, we have for each positive integer $M$:
$$ Pr[ \cup_{n \geq M} \{|X_n-X|>\epsilon\}] \leq \sum_{n=M}^{\infty} Pr[|X_n-X|>\epsilon] $$ It suffices to show the right-hand-side converges to $0$ as $M\rightarrow\infty$. But this is implied by the assumption $\sum_{n=1}^{\infty} Pr[|X_n-X|>\epsilon] < \infty$, since the limit of the tail-sum of a finitely-summable sequence is zero. $\Box$
Proof (Claim 2): Fix $\epsilon>0$. For each positive integer $n$: $$ Pr[|X_n-X|>\epsilon] = Pr[|X_n-X|^d>\epsilon^d] \leq \frac{E[|X_n-X|^d]}{\epsilon^d} $$ where the last inequality is the Markov inequality. Hence: $$ \sum_{n=1}^{\infty} Pr[|X_n-X|>\epsilon] \leq \frac{1}{\epsilon^d}\sum_{n=1}^{\infty} E[|X_n-X|^d] < \infty $$ Then Claim 1 implies $X_n\rightarrow X$ with probability 1. $\Box$