Prove $(X_n, F_n) $ Martingale $\iff \int_{F} X_{n+1} = \int_{F} X_{n} \forall F \in F_n$ I have some additional questions to this exercise:
Let $(\Omega, F, F_n, P)$ filtered probability space. Let $(X_n)_{n\in\mathbb{N}} \in \mathcal{L}^1(P)$, which is adopted to $(F_n)_{n\in\mathbb{N}}$.
Show $(X_n, F_n) $ Martingale $\iff \int_{F} X_{n+1} = \int_{F} X_{n} \forall F \in F_n$. 
I think I have proved the following: "$\Rightarrow$"
$$\int_F X_{n+1} dP = E[1_{F}X_{n+1}] = E[E[1_{F}X_{n+1}|F_n]] = E[1_FE[X_{n+1}|F_n]] = E[1_F X_{n}] = \int_F X_{n} dP $$
Is this correct? Im not sure about this:
$$ E[E[1_{F}X_{n+1}|F_n]] = E[1_FE[X_{n+1}|F_n]] $$
And how should I show the other implication? Any help is appreciated?
 A: The equivalence is just immediate from the definitions of the terms involved. Indeed, with measurability and integrability as background assumptions, we have 
$$(X_n, F_n) \  \text{is a martingale}$$ if and only if 
$$E(X_{n+1} \mid F_n) = X_n$$ if and only if 
$$\int_F X_{n+1} dP = \int_F X_n dP, \ \ \forall F \in F_n,$$
where we used the definition of a martingale first and the definition of conditional expectation second.
As to what you wrote, it's correct, just overkill. The equality you ask about holds because $E(1_FX_{n+1} \mid F_n) = 1_FE(X_{n+1} \mid F_n)$, which is in turn justified by noting that $F \in F_n$ implies that $1_F$ is measurable $F_n$.
A: The two statements $X_n$ is $F_n$-measurable and $\int_{F} X_{n+1} = \int_{F} X_{n} \forall F \in F_n$
are equivalent to
$$E[X_{n+1} | F_n] = X_n$$
provided $X_n$ is integrable, which it is.
Now for $(X_n)_n$ to be an $((F_n)_n, P)_-$ martingale, we need that


*

*$X_n$ is integrable for each $n$ --> given

*$X_n$ is adapted to $F_n$ for each $n$ --> true by definition of $F_n$

*$E[X_{n} | F_m] = X_m$ for $m \le n$ and (I guess?) $m, n \in \mathbb N$


Well, $X_m$ is $F_m$-measurable so all that's left to check is $$\int_{F} X_{m} = \int_{F} X_{n} \forall F \in F_m$$
So yeah, we just have to show that
$$\int_{F} X_{m} = \int_{F} X_{n} \forall F \in F_m \iff \int_{F} X_{n+1} = \int_{F} X_{n} \forall F \in F_n$$
For $\leftarrow$, we know that
$$\int_{F} X_{n+1} = \int_{F} X_{n} \forall F \in F_n \tag{1}$$
and
$$\int_{F} X_{n} = \int_{F} X_{n-1} \forall F \in F_{n-1} \tag{2}$$
Actually, $(1)$ implies
$$\int_{F} X_{n+1} = \int_{F} X_{n} \forall F \in F_{n-1} \tag{3}$$
Thus from $(2)$ and $(3)$ we have
$$\int_{F} X_{n+1} = \int_{F} X_{n-1} \forall F \in F_{n-1} \tag{3}$$
Then we work our way down from $n+1$ to $n$ to $n-1$ all the way until $m$. Formally, I guess we take cases. If $m=n-1$, then we're done. If $m=n-2$, then perform another iteration. And so on. I've never seen an iteration kind of proof in maths before. It just seemed like obvious alternative definitions of the martingale property.
Anyhoo for $\rightarrow$, I think you can take it from here. It's similar to proof for $\leftarrow$, perhaps the steps are reversible. I'm hungry now. So, I'll let you try.
