I apologize in advance if this question is too basic, but I've received no response on Math Stack Exchange, so perhaps it is more appropriate here:


Let $X\_n$ be a square-integrable martingale with $\mathbb{E}\lbrack X\_n^2\rbrack=O(n)$. Is it true that $X\_n/n$ tends to $0$ almost surely?


Note that if one demanded instead that $\mathbb{E}\lbrack X\_n^2\rbrack =O(n^{1-a})$ for some $a>0$, then the claim would follow for any random variables from Markov's Inequality and the Borel-Cantelli Lemma.



Law of large numbers for martingales
The answer is yes, and it is based on an idea by Prokhorov (cf. e.g. Theorem 10 in Section 3 of Ch. IX in [Petrov, V. V., Sums of independent random variables, Springer-Verlag, 1975]). We have $EX\_n^2\le Cn$ for some real $C>0$ and all natural $n$. For natural $s$, let 
\begin{equation}
 T\_s:=\max\_{2^s\le n<2^{s+1}}\frac{|X\_n-X\_{2^s}|}{2^s}. 
\end{equation}
Then, by **[[Doob's martingale inequality](https://en.wikipedia.org/wiki/Doob's_martingale_inequality)]**, for any real $t>0$ 
\begin{equation}
 P(T\_s\ge t)\le\frac{E(X\_{2^{s+1}}-X\_{2^s})^2}{(2^s t)^2}
 \le\frac{EX\_{2^{s+1}}^2}{(2^s t)^2}\le\frac{C2^{s+1}}{(2^s t)^2}=\frac{2C}{2^s t^2}, 
\end{equation}
so that $\sum\_{s=1}^\infty P(T\_s\ge t)<\infty$. So, by the Borel--Cantelli lemma, $T\_s\to0$ almost surely (a.s.) as $s\to\infty$. 
Therefore, for any natural $n$ and $r$ such that $2^r\le n<2^{r+1}$ (so that $r=r\_n:=\lfloor\log\_2 n\rfloor$), one has
\begin{equation}
 \frac{|X\_n-X\_1|}n\le2 \frac{|X\_n-X\_1|}{2^{r+1}}
 \le2\frac1{2^{r+1}}\,\sum\_{s=0}^r 2^s T\_s\to0 \tag{1}
\end{equation}
a.s., since $\sum\_{s=0}^r 2^s<2^{r+1}$; cf. e.g. Lemma 9 in Section 3 of Ch. IX in [Petrov, V. V., Sums of independent random variables, Springer-Verlag, 1975]). Thus, $\frac{X\_n}n\to0$ a.s., as desired. 




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Details on $(1)$: Since $T\_s\to0$ a.s., without loss of generality for each real $\epsilon>0$ there is a natural-valued random variable $R\_\epsilon$ such that for any natural $s$ the event $s>R\_\epsilon$ implies $T\_s\le\epsilon$. Therefore,  

\begin{equation}
 \frac1{2^{r+1}}\,\sum\_{s=0}^r 2^s T\_s
 \le\frac1{2^{r+1}}\,\sum\_{s=0}^{R\_\epsilon} 2^s T\_s+\frac1{2^{r+1}}\,\sum\_{s=R\_\epsilon+1}^r 2^s \epsilon
 \le\frac1{2^{r+1}}\,\sum\_{s=0}^{R\_\epsilon} 2^s T\_s+\epsilon. 
\end{equation}
So, $\limsup\_{r\to\infty}\frac1{2^{r+1}}\,\sum\_{s=0}^r 2^s T\_s\le\epsilon$, for any $\epsilon>0$, and hence $\lim\_{r\to\infty}\frac1{2^{r+1}}\,\sum\_{s=0}^r 2^s T\_s=0$.


