Let $X_n$ be a non-negative decreasing sequence of random variables i.e. $X_n < X_{n-1}$ for $n\in\{1,2,...,\}$ such that $\mathbb P[0\le X_0<\infty] = 1$.
Is $\sup\mathbb E[X_n]<\infty$ ?
Is the supremum finite when $\mathbb E[X_0]<\infty$? $(n\in \mathbb N\cup\{0\})$
I think that I should try to use the monotony of $E$ but I don't know how to start the proof.
If $\{X_n\}_{n=0}^\infty$ are non-negative and decreasing, then monotonicity gives us that:
$\mathbb{E}[X_n]\leq \mathbb{E}[X_0]$ for all $n$, and if $\mathbb{E}[X_0]<\infty$, then $\sup \mathbb{E}[X_n]<\infty$. If however $\mathbb{E}[X_0]=\infty$, I think the following counter-example shows this need not be the case:
$\Omega=[0,1]$ and $X_0(\omega)=2^{m+3}$ for $\omega\in \Big[ \sum\limits_{k=1}^m \frac{1}{2^k},\sum\limits_{k=1}^{m+1} \frac{1}{2^k} \Big]$ with $\mathbb{P}\Bigg( \Big[ \sum\limits_{k=1}^m \frac{1}{2^k},\sum\limits_{k=1}^{m+1} \frac{1}{2^k} \Big]\Bigg)=\frac{1}{2^{m+1}}$. And define: $X_n(\omega)=2^{m+2}+\frac{1}{n}$ for $\omega\in \Big[ \sum\limits_{k=1}^m \frac{1} {2^k},\sum\limits_{k=1}^{m+1} \frac{1}{2^k} \Big]$, with $\mathbb{P}\Bigg( \Big[ \sum\limits_{k=1}^m \frac{1}{2^k},\sum\limits_{k=1}^{m+1} \frac{1}{2^k} \Big]\Bigg)=\frac{1}{2^{m+1}}$. Then they all have infinite expectation, and $\sup \mathbb{E}[X_n]=\infty$.