Undestanding the Martingale Representation Theorem 
Theorem (Representation Theorem):Let $X$ be a binary model and let $V_T$ be an $\mathscr{F}_T$-measurable random variable. Then there exists a bounded predictable process $H$ and a $v_0\in\mathbb{R}$ with $V_T=v_0+(H.X)_T$.

Proof: We show that there exists $\mathscr{F}_{T-1}$-measurable random variables $V_{T-1}$ and $H_T$ such that $V_T=V_{T-1}+H_T(X_T-X_{T-1})$. By a backward induction, this yields the claim.
Since $V_T$ is $\mathscr{F}_T$-measurable, by the factorisation lemma, there exists a function $g_{T}:\mathbb{R}^T\to\mathbb{R}$ with $V_T=g_T(X_1,...,X_{T})$. Define:
$X^{\pm}_T=f_T(X_1,...,X_{T-1},\pm 1)$ and $V^{\pm}_T=g_T(X_1,...,X_{T-1},X^{\pm}_T)$.
Each of these four random variables is manifestly $\mathscr{F}_{T-1}$-measurable. Hence we are looking for solutions $V_{T-1}$ and $H_T$ of the following system of linear equations:
$V_{T-1}+H_T(X^{-}_T-X_{T-1})=V^{-}_T\\V_{T-1}+H_T(X^{+}_T-X_{T-1})=V^{+}_T$
By construction, $X^+_{T}-X^{-}_T\neq 0$ if $V^{+}_T-V^{-}_T\neq 0$. Hence we can solve the system and get:
$H_t=\begin{cases}
\frac{V^{+}_T-V^{-}_T}{X^+_{T}-X^{-}_T}, & \mbox{ if }X^+_{T}\neq X^{-}_TE\\
0, & \mbox{ else }\end{cases},$
and 
$V_{T-1}=V^{+}_T-H_T(X^{+}_T-X_{T-1})=V^{-}_T-H_T(X^{-}_T-X_{T-1})\:\:\:\:\:\:\:\blacksquare$
I am trying to understand this proof but I do not know if I understood it right. 
Questions:
1) When it starts proving by backward induction. Does it mean that $V_T=V_{T-1}+H_T(X_T-X_{T-1})$  is assumed to be true? 
2) Why is  $X^+_{T}-X^{-}_T\neq 0$ if $V^{+}_T-V^{-}_T\neq 0$ true? Why is not the other way around?
3)Why is the proof complete after solving the system and obtain $H_T$? 
Thanks in advance!
 A: I provide directions to answer your questions:


*

*$\forall t$ in 0,...,T$, V_t=V_{t-1}+H_T(X_t-X_{t-1}) \\
\Rightarrow \sum\limits_{t=1}^T V_t-V_{t-1}=\sum\limits_{t=1}^TH_t(X_t-X_{t-1}) \\ \Rightarrow V_T-V_0=\sum\limits_{t=1}^TH_tX_t - \sum\limits_{t=0}^{T-1}H_{t+1}X_{t} $
$H^1_t = ? $ which is also predicable
$v_0 = V_0$ 
The first line is not assumed true. It is proven. The author starts stating that proving it is enough to prove the theorem. Here again he uses a fancy wording. The proof of the statement $V_t=V_{t-1}+H_T(X_t-X_{t-1})$ starts at the word "Since $V_T$ is..". 


*It comes from this equation ($*$) $V_{T-1}+H_T(X^{-}_T-X_{T-1})=V^{-}_T$. The author assume this true to search for a candidate H. He finds the candidate H which proves ($*$). It is an Analysis-Synthesis Reasoning.
he built $X^+$, $X^-$, $V^-$ and $V^+$ (which are $\mathscr{F}_{T-1}$ measurable because they only depend on $X_{T-1},...,X_1$) in order to ultimately construct a H process which is predictable.

*First, notice the following relationship: $V_T = g_T(X_1,..,X_T) = g_T(X_1,..,X_{T-1},1) 1_{X_T=1}+ g_T(X_1,..,X_{T-1},-1) 1_{X_T=-1} \\ =  V_T^+ 1_T^+ + V_T^- 1_T^- \\
\text{Likewise, }  X_T=X_T^+ 1_T^+ + X_T^- 1_T^-$
We have
 ($E^+$) $V_{T-1}+H_T(X^{+}_T-X_{T-1})=V^{+}_T$ and ($E^-$) $V_{T-1}+H_T(X^{-}_T-X_{T-1})=V^{-}_T$
Let's do $(E^+) \times 1^+_T + (E^-) \times 1^-_T$
It proves $V_{T-1} =V_T - H_T(X_T)*X_{T-1} $ 
I add the following intuitions for less experienced readers:
Filtration:
In probability a filtration $\mathscr{F}_T$ (which is "generated" by $(X_T)$ in this context) can be thought of the mathematical object gathering all the information available after observing ($X_T$) during time [0,T] (as if you were storing all the information associated to X in the world at time T in $\mathscr{F}_T$). It contains more information that just the realized values. Any function dependent on the joint distribution $(X_1,..X_T)$ is known conditionally to $\mathscr{F}_T$.
Measurability:
You can think of being $\mathscr{F}_T-measurable$ as being entirely determined (deterministic meaning that you can get the value), by the observation of the past distribution of X up to T (once again not only the variable $X_t$ and their realized values $x_t$ but the joint distribution). In simple mathematical terms, it can be characterized by the factorization lemma: $V_T =g_T (X_1 ,...,X_T )$ 
