Exercise 4.3.3 (A first look at rigorous probability theory) The exercise aims to prove linearity of expectation for general (i.e. not necessarily non-negative) random variables. It is defined that for a general r.v., $E[X] = E[X^+] -E[X^-]$, where $X^+$ is the positive part of $X$, $X^-$ is the magnitude of the negative part of $X$. In particular, $X^+,X^-$ are non-negative random variables.
Let $X$ and $Y$ be two general random variables with finite means, and let $Z = X + Y$.

*

*(a) Express $Z^+-Z^-$ in terms of $X^+,X^-,Y^+,Y^-$

*(b) Prove that $E[Z^+]-E[Z^-]=E[X^+]-E[X^-]+E[Y^+]-E[Y^-]$
I am stuck at the second part. I get $E[Z] = E[X^+-X^-+Y^+-Y^-] = E[(X^++Y^+) - (X^-+Y^-)]\stackrel{?}{=}E[X^++Y^+]-E[X^-+Y^-]=E[X^+]-E[X^-]+E[Y^+]-E[Y^-]$.
There is no direct result in the book that shows $\stackrel{?}{=}$ to hold.
Because for non-negative random variables $X,Y$, it is only proven that $\forall a,b\geq 0, E[aX+bY] = aE[X]+bE[Y]$. Additionally, the book's exercise claims that this property is sufficient to do this exercise. (It literally says "Hint: Re-arrange the relations of part (a) so
that you can make use of (4.2.6).")

You are close. You will be done as long as you can justify
$$
E[Z]= E[(X^++Y^+) - (X^-+Y^-)]\stackrel{?}{=}E[X^++Y^+]-E[X^-+Y^-]\tag{$\star$}
$$
To prove this equation, start by writing
$$
\begin{align}
Z &= X+ Y,\\
Z^+-Z^-&=(X^+-X^-)+(Y^+-Y^-)
\end{align}
$$
Rearrange this to
$$
Z^++X^-+Y^-=Z^-+X^++Y^+
$$
Now, all summands are nonnegative. Therefore, we can take $E[\,\cdot\,]$ of both sides and distribute the $E$:
$$
E[Z^+]+E[X^-]+E[Y^-]=E[Z^-]+E[X^+]+E[Y^+]
$$
You now have all you need to prove ($\star$).
