probability almost surely and expectation It is true that $P(X= Y)=1 \Rightarrow E(X)= E(Y)$ and $P(X\ge Y)=1 \Rightarrow E(X)\ge E(Y)$, but is it true that $P(X > Y)=1 \Rightarrow E(X) > E(Y)$ ?
The proofs for the first two don't quite carry over so I'm not sure if this is true or not.
 A: Let $P\{Z>0\}=1$ and realize that $\{Z>0\}=\bigcup_{n=1}^{\infty}\{Z\geq\frac1{n}\}$. 
That implies that $P\{Z\geq\frac1{n}\}$ converges to $P\{Z>0\}=1$ so that $P\{Z\geq\frac1{n}\}>\frac12$ for some $n$. 
Consequently $\mathbb EZ\geq\frac1{n}\frac12>0$.
Apply this on $Z=X-Y$.

Edit:
Alternatively $1=P\left\{ Z>0\right\} \leq\sum_{n=1}^{\infty}P\left\{ Z\geq\frac{1}{n}\right\} $
by subadditivity. 
Then $P\left\{ Z\geq\frac{1}{n}\right\} >0$ for some $n$ implying
the existence of an $\epsilon$ with $P\left\{ Z\geq\frac{1}{n}\right\} \geq\epsilon>0$. 
Then $Z\geq\frac{1}{n}1_{\left\{ Z\geq\frac{1}{n}\right\} }$ a.s.
so that $\mathbb{E}Z\geq\mathbb{E}\frac{1}{n}1_{\left\{ Z\geq\frac{1}{n}\right\} }=\frac{1}{n}\mathbb{P}\left\{ Z\geq\frac{1}{n}\right\} \geq\frac{1}{n}\epsilon>0$.
A: Yes. The proof for the general case is a bit messier than the other two, because as you know we usually only retain nonstrict inequalities under a limit process. The idea is that the set $X>Y$ can be broken up into a union of sets $X>Y+1/n$. If any of these has positive probability, and $X<Y$ has zero probability, then $E[X]>E[Y]$ (why?). The way we show that one of them has positive probability is using continuity of measure. We write:
$$\{ X>Y \}=\bigcup_{n=1}^\infty \{ X > Y + 1/n \}.$$
Now the union on the right is an increasing union, so we get
$$P(X>Y) = \lim_{n \to \infty} P(X>Y+1/n).$$
The definition of limit then implies that we have some $n$ with $P(X>Y+1/n)>1/2$, say. That finishes the proof. (Note that I have used the shorthand notation which is common in probability theory.)
The slightly tricky part is proving continuity of measure; this can be found in essentially any measure theory text, but let me know if you need help with it.
