An example of a martingale converge almost surely but not in $L^1$? Q1 An example of a martingale converge almost surely but not in $L^1$?
This I saw example $P(X_{n+1}=j|X_n=i)=\frac{1}{2i+1}$ where $0\le j\le 2i$ and $X_0=10$ but how to actually prove it?
Q2 An example of a martingale both converge almost surely and also in $L^1$? a stopped coin tossing game is an example I think.
Additionally if you think these are important,
An example of a martingale converge in probability but not in $L^1$?
An example of a martingale converge in probability but not in a.s.?
 A: 
martingale which converges almost surely but not in $L^1$

Let $(M_n)_{n \in \mathbb{N}}$ be a simple random walk and define $$T := \inf\{n \in \mathbb{N}; M_n = 1\}.$$ If we consider the  stopped martingale $N_n := M_{T \wedge n}$, then $N_n \xrightarrow[]{n \to \infty} N_{\infty} := M_T=1$ almost surely. On the other hand, $\mathbb{E}(N_n) = 0 \neq 1= \mathbb{E}(N_{\infty})$, and therefore $(N_n)_{n \in \mathbb{N}}$ does not converge in $L^1$.

martingale which converges both almost surely and in $L^1$

Well, the easiest example is the trivial one, i.e. just set $M_n := X$ for some random variable $X \in L^1$. Alternatively, consider again a simple random walk $(M_n)_{n \in \mathbb{N}}$ and define $$T := \inf\{n \in \mathbb{N}; |M_n| = 1\}.$$ Then the martingale $N_n := M_{T \wedge n}$ is bounded in $L^2$, and this implies $L^2$-convergence (hence $L^1$-convergence) as well as pointwise convergence.

martingale which converges in probability but not in $L^1$

Any martingake which converges almost surely but not in $L^1$ does the job; see the first part of my answer.

martingale which converges in probability, but not almost surely

There are martingales which converge in probability but not almost surely; however, it is not easy to find such martingales. See for instance this article or the book by Stoyanov for (counter)examples.
