Mentioning the definition of Supremum offered in supremum and limsup of random variables , we have: For each $\omega \in \Omega$, $$(\sup_{n \in \mathbb{N}} X_n)(\omega) := \sup\{X_n(\omega):n \in \mathbb{N}\}$$
I tried to imagine an expressive example:
Event: throw a die.
$(X_n)=\frac1n$ when $\omega=1,2$;
$(X_n)=\frac1{2}$ when $\omega=3,4$;
$(X_n)=0$ when $\omega=5,6$.
Then we can compute $$E[X_n]:\frac1n\frac13+\frac1{2}\frac13+0\frac13=\frac3{6}=\frac1{2}$$
Therefore: $\sup_n E[X_n]=\frac12$. For computing $E[\sup X_n]$ instead we have:
$\sup_n X_n(\omega)=1$ if $\omega =1$ or $2$,
$\sup_n X_n(\omega)=\frac 1 2$ if $\omega =3$ or $4$
and $0$ if $\omega =5$ or $6$
Therefore $E[\sup X_n]=1\frac13+\frac12\frac13+0\frac13 =\frac36=\frac12$
Could someone tell me if this reasoning is it correct?