Bell Inequalities - Expectation Values I'm currently reading Loopholes in Bell Inequality Tests of Local Realism by Jan-Ake Larsson
https://arxiv.org/abs/1407.0363
On Page 6, Equation 7, he has a short proof, where I am having a hard time seeing through the math.  I'll re-state here for convenience:
$ \lvert E(A_{2}B_{1}) - E(A_{2}B_{2}) \rvert = \lvert E(A_{2}B_{1} + A_{2}B_{1}A_{1}B_{2}) \rvert \leq E(\lvert A_{2}B_{1}(1 + A_{1}B_{2}) \rvert ) = 1 + E(A_{1}B_{2}) $
where 
$ A_{1}B_{1} = -1 $
and
$E(A) = \int A( \lambda ) \rho ( \lambda) d \lambda $
I'm not seeing how he goes from the 2nd expression on the LHS of the less-than-or-equal sign to the third expression on the RHS of the less-than-or-equal sign.  I suspect I am missing something on the properties of expectation values.  Mainly, I think the absolute signs (|) moving from outside of the E()'s to the inside of the E()'s have me the most confused.  Further, I don't see how he goes from the 3rd expression to the 4th, either.
Can anyone offer any clarification here?
 A: From 2nd to 3rd expression:
It is well known that:
$$\left|\int f(x)dx\right| \leq \int \left|f(x)\right| dx.$$
Here you can find the reference for this fact.
Now, you have that:
$$\lvert E(A_{2}B_{1} + A_{2}B_{1}A_{1}B_{2}) \rvert = \left|\int (A_{2}B_{1} + A_{2}B_{1}A_{1}B_{2})\rho(\lambda)d\lambda \right| \leq \\
\leq \int \left|(A_{2}B_{1} + A_{2}B_{1}A_{1}B_{2})\rho(\lambda)\right|d\lambda = \int \left|A_{2}B_{1}(1 + A_{1}B_{2})\rho(\lambda)\right|d\lambda.$$
Since $\rho(\lambda) \geq 0$ by definition (it is a distribution), then $\rho(\lambda) = |\rho(\lambda)|$, and hence:
$$\int \left|A_{2}B_{1}(1 + A_{1}B_{2})\rho(\lambda)\right|d\lambda = \\
\int \left|A_{2}B_{1}(1 + A_{1}B_{2})\right|\rho(\lambda)d\lambda = E(|A_{2}B_{1}(1 + A_{1}B_{2})|).$$
From 3rd to 4th expression:
Suppose that $B_1=1$. Then:
$$E(|A_{2}B_{1}(1 + A_{1}B_{2})|) = E(|A_{2}(1 + A_{1}B_{2})|) = E(|A_{2}||1 + A_{1}B_{2}|).$$
Since $A_2 \in \{-1, +1\}$, then $|A_2| = 1$, and hence:
$$E(|A_{2}B_{1}(1 + A_{1}B_{2})|) = E(|1 + A_{1}B_{2}|).$$
Since $B_1 = 1$, then $A_1 = -1$, and the term $1 + A_{1}B_{2}$ can be equal to $0$ or to $2$. In each case, it is positive, so we can drop the modulus:
$$E(|A_{2}B_{1}(1 + A_{1}B_{2})|) = E(1 + A_{1}B_{2}).$$
Last steps are easy...
$$E(|A_{2}B_{1}(1 + A_{1}B_{2})|) = E(1 + A_{1}B_{2}) = E(1) + E(A_1 B_2) = 1 +E(A_1 B_2).$$
Now suppose that $B_1 = -1$. Then:
$$E(|-A_{2}(1 + A_{1}B_{2})|) = E(|A_{2}(1 + A_{1}B_{2})|).$$
Using the same reasonings of above, we get that:
$$E(|-A_{2}(1 + A_{1}B_{2})|) = E(|1 + A_{1}B_{2}|).$$
$A_1$ is $1$, and $1 + A_{1}B_{2}$ can be $0$ or $2$. We can drop the modulus... etc... etc... Even in this case, we get that $E(|A_{2}B_{1}(1 + A_{1}B_{2})|) = 1 + E(A_{1}B_{2})$.
A: Let's write $X$ for $A_2B_1$ and $Y$ for $A_2B_1A_1B_2$.  And keep in mind that $X$ and $Y$ are each always equal to either $1$ or $-1$.
Now think of the expectation  as an average. You've got a bunch of possible outcomes, and $E(X+Y)$, for example, is the average value of $X+Y$ over all those outcomes.  To understand how these averages behave, it's enough to understand how the totals of the outcomes behave, since the average is just the total divided by the number of outcomes.
So write $T(X+Y)$ for the total of all the possible outcomes of $X+Y$.  Those outcomes come in three types:  Either $X=Y=1$, which contributes $2$ to the total, or $X=Y=-1$, which contributes $-2$, or $X=-Y$, which contributes $0$.  If there are $m$ observations of the first kind, $n$ of the second, and $k$ of the third, then $T(X+Y)=2m-2n$, and $|T(X+Y)|=|2m-2n|$ (which is either equal to  $2m-2n$ or $2n-2m$, whichever is positive).
Now $T(|X+Y|)$ is the same total, except that all of the $-2$'s have been converted to $2$'s.  $So |T(X,Y)|=2m+2n$.
This makes it clear that $|T(X+Y)|\le T(|X+Y)|)$.  If that's not already crystal clear,  think of it this  way:  The left hand side involves some positive and negative terms that cancel, while on the right hand side all the negatives have been converted to positives, so no cancellation occurs.  
To pass from $T$ to $E$, just divide by the number of possible outcomes.  That tells you that $|E(X+Y)|\le E(|X+Y|)$, which is the inequality between the second and third expressions.
For the final equality, note that $A_1B_2$ is always either $1$ or $-1$, so multiplying by it doesn't change the absolute value of anything.  
(This is all a special case of the more general theorem that Candyman quotes in his answer, but the special case is all you need and might  be easier to grasp. On the other hand, it's well worth mastering the general theorem at some point.)
Now for the final step:
First, $A_2B_1$ is always either $1$ or $-1$.  Either way, if $x$ is anything at all, $A_2B_1x$ is going to have the same absolute value as $x$.  If we take $x=1+A_1B_2$, we get $E(|A_2B_1(1+A_1B_2)|)=E(|1+A_1B_2|)$. 
Second, $A_1B_2$ is always either $1$ or $-1$.  Either way, $1+A_1B_2$ is non-negative, and hence equal to its own absolute value.  This lets you drop the absolute value signs.
