Positivity and Strict Positivity of Conditional Expectations

Question

I am reading this set of probability theory notes from Stanford:

http://statweb.stanford.edu/~adembo/stat-310b/lnotes.pdf

And I was wondering if someone could help me understand his proof of proposition 4.2.4. page 157.

Question 1:

"Hence in this case: \begin{equation*} E\left[ Y I_{Y\le 0}\right] = 0 \end{equation*} That is, $Y \ge 0$ almost surely."

I cannot see for the life of me why this is the case....I know the fact:

If A is a measurable set with measure zero, i.e. $\mu(A) = 0$, then the integral of a corresponding measurable random variable over this set is equal to 0. But to my knowledge I am not aware that the converse of this statement is true!

Question 2: Also when he writes: "...so $P(X>0, Y = 0) = 0$...should it instead be that $P(X>0, Y \le 0)$?

I am also quite uncertain as to how this last statement implies that $Y> 0$ a.s.

My gut tells me that if the above "corrected" statement was true, then because $P(X>0) = 1$, this must imply that that $P(Y\le 0) = 0$...

Thanks for the help in advanced!

Answer 1

Question 1: The point is that if $P(Y < 0) > 0$, then $$0 \leq E(XI_{Y<0}) = E(YI_{Y < 0}) < 0$$ is a contradiction. Hence, we must have $P(Y \geq 0)=1$ whenever $P(X \geq 0)=1$.

Question 2: Note that $X > 0$ a.s. implies $Y \geq 0$ a.s. by part 1. So to show that $Y > 0$ a.s. it remains to show that $P(Y=0)=0$. Since $P(X > 0)=1$, then by what you've written $0 = P(X > 0, Y=0) = P(Y=0)$.

Answer 2

Notice that $Y I_{Y \le 0}$ must be non-positive almost surely. The fact that $E\left[Y I_{Y \le 0}\right]=0$ then implies that $Y I_{Y \le 0} = 0$ (a.s.). You can prove this by the standard machine. (Alternatively, see theorem 1.3.9(c).) This is enough to conclude what you need for the first question.

For your second question, note that you have already proven that $Y \ge 0$ from the earlier part of the proof, so you can write $Y=0$ instead of $Y \le 0$.