Proving a sequence forms a martingale Let $\Omega = \mathbb N = \{1,2,3,\cdots\}$ and $\mathscr F_n$ be the $\sigma$-field generated by the sets $\{1\},\{2\},\cdots,\{[n+1,\infty)\}$
Define a probability on $\mathbb N$ by setting $\mathbb P([n,\infty)) = \frac 1 n$
Show that 
1) $f_n =(n+1)\mathbf1_{[n+1,\infty)}$ is a martingale
2) $f_n \to 0 \quad a.s.$ 
I'm having difficulty dealing with the indicator r.v. How do we get it out of the conditional expectation for $\mathbf E[f_{n+1}\mid \mathcal{F}_n]$?
I tried to graph it out and I think I can see why it forms a martingale but I'm not able to prove it explicitly.
$f_n$">
 A: Hints:


*

*It is well-known that a random variable $Y \in L^1$ equals $\mathbb{E}(X \mid \mathcal{F})$ if and only if it is $\mathcal{F}$-measurable and $$\int_G Y \, d\mathbb{P} = \int_G X \, d\mathbb{P} \qquad \text{for all $G \in \mathcal{G}$}$$  where $\mathcal{G}$ denotes  a $\cap$-stable generator of $\mathcal{F}$, i.e. $\sigma(\mathcal{G}) = \mathcal{F}$ and $$G_1, G_2 \in \mathcal{G} \Rightarrow G_1 \cap G_2 \in \mathcal{G}.$$ Check that $\mathcal{G}_n := \{\emptyset,\{1\},\{2\},\ldots,[n+1,\infty)\}$ is a $\cap$-stable generator of $\mathcal{F}_n$.

*In order to show that $f_n$  is a martingale, it suffices to show $$\mathbb{E}(f_n \mid \mathcal{F}_{n-1}) = f_{n-1};$$ by step 1 this is equivalent to $$\int_G f_n \, d\mathbb{P} = \int_G f_{n-1} \, d\mathbb{P} \tag{1}$$ for all $G \in \mathcal{G}_{n-1}$.

*By the very definition of $f_n$, we have $$\begin{align*} A&:= \int_G f_n \, d\mathbb{P} = (n+1) \mathbb{P}(G \cap [n+1,\infty)) \\  B &:= \int_G f_{n-1} \, d\mathbb{P} = n \mathbb{P}(G \cap [n,\infty)). \end{align*}$$ We have to prove $A=B$ for all $G \in \mathcal{G}_{n-1}$. Consider two cases separately:

*

*$G = \{j\}$ for some $j \in \{1,\ldots,n-1\}$: Since $G \cap [n\infty) = G \cap [n+1,\infty)=\emptyset$, we get $A=B=0$.

*$G = [n,\infty)$: Conclude from $G \cap [n,\infty)= [n,\infty)$ and $G \cap [n+1,\infty) = [n+1,\infty)$ that $A=B=1$.


