One property of random variables is that there exists $\omega\in\Omega$ such that $X(\omega)\ge\Bbb E(X)$ (if someone could point me to a proof of that I'd be thankful).
But there are finite variable with infinite expectation, so I was wondering what goes wrong in the above theorem because obviously we cannot have that $X(\omega)\ge\infty$
If no such $\omega$ exists then $Y$ prescribed by $\omega\mapsto X(\omega)-\mathbb EX$ is a negative random variable.
Then on base of $P(\bigcup_{n=1}^{\infty}\{Y<-\frac1n\})=P(\Omega)=1$ it can be shown that some $n\in\mathbb N$ exist such that $P(Y<-\frac1n)>0$ which leads in combination with the fact that $Y$ is negative to the conclusion that $\mathbb EY<0$.
This however contradicts that $\mathbb EY=\mathbb E(X-\mathbb EX)=\mathbb EX-\mathbb EX=0$.
This reasoning does not work in the case of infinite expectation. Then we cannot subtract $\mathbb EX$ as we did above.