How do I conclude from $E[X_{+}1_{Y\geq 0}]=0$ that $P(X > 0, Y \leq 0)=0$ Let $X_{+}$ be the part of $X$ that is strictly positive. 
How do I conclude from $E[X_{+}1_{Y\geq 0}]=0$ that $P(X > 0, Y \leq 0)=0$?
My idea
$0=E[X_{+}1_{Y\geq 0}]=E[X1_{X>0,Y\leq0}]$ and since $X\neq 0$ on $1_{X>0}$ it follows that $P(X>0,Y\leq 0)$
it seems obvious but I do not think my reasoning is up to scratch. Any ideas? Is strictly positive actually necessary here?

Your claim is wrong. Consider $X$ a random variable with exponential distribution and let $Y=-X$. Then $X_{+}1_{Y\geq 0} = 0$ a.s., thus $E(X_{+}1_{Y\geq 0} )=0$. However $$P(X>0\cap Y\leq 0) = P(X>0)=1 $$
The only thing you can derive from $E(X_{+}1_{Y\geq 0} )=0$ is $X_{+}1_{Y\geq 0}$ a.s., which is equivalent to $(X>0 \implies Y<0)$ a.s. Hence 
$$\begin{aligned}
P(X>0\cap Y\leq 0) &= E(1_{X>0}1_{Y\leq 0})\\
&= E(1_{X>0}1_{Y< 0}) + E(1_{X>0}1_{Y= 0})\\
&= E(1_{X>0}) \\
&=P(X>0)
\end{aligned}$$
