I will use the following reference: Klenke
Unfortunately the book is in German, nevertheless here is my answer, based on it:
On p. 208 Klenke shows ("Satz 10.4) that for a square integrable martingale $\{X\}$ the so called quadratic variation process is given by
$ \langle X \rangle_n = \sum_{i=1}^n E((X_i-X_{i-1})^2|\mathcal{F}_{i-1}).$
Since you have already proved, that $ M_n$ is a square integrable martingale, we find:
$ \langle M \rangle_n = \sum_{i=1}^n E((M_i-M_{i-1})^2|\mathcal{F}_{i-1})$
since $M_n$ is defined through a sum, this is equal
$\langle M \rangle_n = \sum_{i=1}^n E((X_i-E(X_i|\mathcal{F}_{i-1}))^2|\mathcal{F}_{i-1}).$
So far nothing happens, but it's important for the things, which follow:
The main Theorem can be found on page 225, "Korollar 11.11", I cite (and translate):
If $ X = \{X_n\}$ is a square integrable martingale with quadratic variation process $\langle X \rangle $ the following are equivalent:
- $\sup_n E(X_n^2) < \infty $
- $\lim_n E(\langle X\rangle_n) < \infty$
- $X$converges in $ L^2$.
- $X$ converges in $ L^2$ and almost surely.
What we will use is the equivalence of $2.\iff 4.$ First let me point out, that I do not see how to prove your 3. without knowing that $ M_\infty \le c < \infty$. Though it's not hard to prove your 4.
Assume that $ \sum_{i=1}^\infty E(X_i^2) < \infty $ then we want to show, that $M_n$ converges a.s.
As mentioned, I will use the equivalence of $2.\iff 4.$: $\lim E(\langle M\rangle_n) = E(\langle M\rangle_\infty)$ using monotone convergence once more,
$E(\langle M\rangle_\infty)=\sum_{i=1}^\infty E(E((X_i-E(X_i|\mathcal{F}_{i-1}))^2|\mathcal{F}_{i-1}))$
by basic properties of conditional expectation:
$\sum_{i=1}^\infty E(E((X_i-E(X_i|\mathcal{F}_{i-1}))^2|\mathcal{F}_{i-1})) =\sum_{i=1}^\infty E(E((X_i-E(X_i|\mathcal{F}_{i-1}))^2).$
Now observe:
$0\le(a-b)^2=a^2-2ab+b^2=|a^2-2ab+b^2|\le a^2+2|ab|+b^2$
and by basic analysis, $ 2|ab| \le a^2 +b^2 \Rightarrow (a-b)^2\le 2a^2+2b^2$, this leads to:
$E(E((X_i-E(X_i|\mathcal{F}_{i-1}))^2)\le 2(E(X_i^2) + E(E(X_i|\mathcal{F}_{i-1})^2)).$
Now use Jensen for conditional expectation $ E(E(X_i|\mathcal{F}_{i-1})^2) \le E(E(X_i^2|\mathcal{F}_{i-1}))= E(X_i^2)$, hence
$ \sum_{i=1}^\infty E(E((X_i-E(X_i|\mathcal{F}_{i-1}))^2)\le 4\sum_{i=1}^\infty E(X_i^2) < \infty$
by assumption, and therefore $ M_n $ converges a.e. and even more, we see that $ \langle M \rangle_\infty < \infty$ a.s.
As I said, I don't know how to prove 3. but I decided to post an answer, because maybe my calculation helps someone to prove 3. and second, it's too long for a comment.
cheers
math