Let $X_1,X_2,\dots$ be a sequence of equally distributed random variables.
Suppose that $\forall n\in\mathbb{N}:\mathbb{E}[X_n]=0$ and $\lim\limits_{n\to\infty}\mathbb{E}[X_n^4]=0$.
Prove or Disprove: $X_n\stackrel{\text{a.s.}}\to0$.
I feel that the statement is true and I tried to prove it.
I said that since the random variables are all equally distributed then the random variables $|X_1|,|X_2|,\dots$ are also equally distributed.
So, $\forall n,m\in\mathbb{N}:\mathbb{E}[|X_n|]=\mathbb{E}[|X_m|]$. Let's call this value $S$.
Since $\lim\limits_{n\to\infty}\mathbb{E}[X_n^4]=0$, it must be that $S=0$ (not sure about this).
Therefore, $P[X_n=0]=1$ and $X_n\stackrel{\text{a.s.}}\to0$.
Is it true?
As mentioned in the comments we have $E[X_n^4]=0$, i.e. \begin{align} \int_\Omega |X_n|^4 \,dP = 0 \end{align} And that means $X_n=0$ a.s.