skorokhod embedding from Rogers Williams I am reading Rogers and Williams "Diffusions, Markov Processes and Martingales" and have a question about the proof of the Skorohod embeding. What we want to proof:

Let $X$ be a zero-mean finite variance $\sigma^2$. r.v. with distribution $F$. We want to find a stopping time $T$ such that $B_T\sim F$ and $ET=\sigma^2$.

$B$ denotes a Brownian Motion. First thing the authors do is to define the distribution $\mu (da,db):=\gamma (b-a)F_+(db)F_-(da)$, where $F_+,F_-$ are the restriction on $[0,\infty), (-\infty,0)$ of $F$ and $\gamma$ a normalization constant. I know that for $\tau:=\inf\{t:B_t\notin (a,b)\}$, where $a<0<b$ has the property
$$P(B_\tau=b)=\frac{-a}{b-a}$$
$$E\tau=|ab|$$
What is meant by: pick a pair $\alpha <0<\beta$ according to the distribution $\mu$? Is it meant that I choose $\alpha$ according to $F_-$ and $\beta$ to $F_+$? 
The stopping time $T:=\inf\{u:B_u\notin (\alpha,\beta)\}$ should do the job. The following equations are not clear to me:
for $b\ge0$:
$$P(B_T\in db)=\int_{(-\infty,0)}\frac{-a}{b-a}\gamma F_+(db)F_-(da)$$
so we are integrating over $da$? Moreover, why is 
$$ET=\int_{[0,\infty)}\int_{(-\infty,0)}\mu(da,db)|ab|$$ and why is
$$\gamma\int_{[0,\infty)}F_+(db)\int_{(-\infty,0)}F_-(da)(b-a)|ba|=\int_{-\infty}^{\infty} x^2 F(dx)$$
Thanks in advance
 A: Wow, in fact you do not have "a question about the proof of the Skorohod embed(d)ing", you have an awful lot of them.

What is meant by: pick a pair $\alpha <0<\beta$ according to the distribution $\mu$? Is it meant that I choose $\alpha$ according to $F_-$ and $\beta$ to $F_+$? 

No, one picks a random vector $(\alpha,\beta)$ whose joint distribution is $\mu$. What you suggest amounts to choose a random vector with joint distribution (a multiple of) the product of $F_+$ and $F_-$, which $\mu$ is not (except in specific degenerate cases) because of the factor $b-a$.

The following equations are not clear to me

None of them?
Consider the very last identity. The LHS is the sum of two terms, the first one being
$$
\gamma\int_{[0,\infty)}F_+(db)\int_{(-\infty,0)}F_-(da)b|ba|=\gamma\int_0^{+\infty}b^2F_+(db)\int_{-\infty}^0|a|F_-(da),
$$
that is, $A$ times the RHS,
for some explicit constant $A$. Likewise the second term is $B$ times the RHS and all you have to check is that $A=B=1$. Hint: here the condition that $\mathbb E(X)=0$ is crucial, rewritten as $\mathbb E(X;X\geqslant0)=\mathbb E(|X|;X\lt0)$, that is,
$$
\int_0^{+\infty}bF_+(db)=\int_{-\infty}^0|a|F_-(da).
$$
Edit: The definition of $T$ is as follows. For every $a\lt0\lt b$, let $\tau(a,b):=\inf\{t:B_t\notin (a,b)\}$, hence $(\tau(a,b))_{a,b}$ depends on $(B_t)_t$ only and, for every $(a,b)$, $\mathbb E(\tau(a,b))=\theta(a,b)$ where $\theta(a,b)=|ba|$. Independently of $(B_t)_t$, draw $(\alpha,\beta)$ according to the distribution $\mu$, then
$$
T=\tau(\alpha,\beta).
$$
In particular,
$\mathbb E(T\mid\alpha,\beta)=\theta(\alpha,\beta)$ and
$$
\mathbb E(T)=\mathbb E(\mathbb E(T\mid\alpha,\beta))=\mathbb E(\theta(\alpha,\beta))=\mathbb E(|\alpha\beta|)=\iint |ba|\mu(\mathrm da,\mathrm db).
$$
Likewise, using the shorthand $W(a,b)=B_{\tau(a,b)}$, one gets
$$
B_T=W(\alpha,\beta),
$$
hence, for every Borel subset $J\subseteq(0,+\infty)$,
$\mathbb P(B_T\in J\mid\alpha,\beta)=Q_J(\alpha,\beta)$ where
$Q_J(a,b)=\mathbb P(W(a,b)=b)\mathbf 1_{b\in J}$, hence
$$
\mathbb P(B_T\in J)=\mathbb E(\mathbb P(B_T\in J\mid\alpha,\beta))=\iint\mathbb Q_J(a,b)\mu(\mathrm da,\mathrm db),
$$
which yields
$$
\mathbb P(B_T\in \mathrm db)=\int\mathbb P(W(a,b)=b)\mu(\mathrm da,\mathrm db),
$$
where the integral in the RHS is over $a$ only, that is,
$$
\mathbb P(B_T\in \mathrm db)=\int_{-\infty}^0\frac{-a}{b-a}\mu(\mathrm da,\mathrm db).
$$
