Suppose I have a sequence of integrable random variables $(X_n)$. If I define another random variable $Y:=\sup_n\{(X_n)\}$, then is $Y$ also integrable? I'm asking this question because, if true, it can help me solve a problem that I'm working right now:
Suppose $(X_n)\rightarrow X$ in probability or almost everywhere. Suppose too that $|X_n|\leq Y_n$, $Y_n\rightarrow Y$ almost everywhere, and $\mathbb{E}(Y_n)\rightarrow\mathbb{E}(Y)$. $Y_n$, $Y$, $X_n$, and $X$ are all integrable. Show that $\mathbb{E}(X_n)\rightarrow\mathbb{E}(X)$.
My hope was to use the dominated convergence theorem by comparing $(X_n)$ to $\sup_n\{(Y_n)\}$. Is this correct, or is there some other method of solving this problem?
It's not true. For a trivial example, consider $X_n:=n$ for each $n$. A less trivial example would be to define on the unit interval $$X_n(\omega):=nI\left(\omega\le \frac1n\right)=nI_{[0,\frac1n]}(\omega).$$ Then each $X_n$ is a rectangle of width $\frac1n$ and height $n$, hence has integral $1$. You can confirm that $\sup X_n\ge Z$ where $Z$ is defined by $$Z(\omega):=\frac1\omega-1,$$ and $Z$ is not integrable.
As for the result you are trying to prove: One way is to first show that the sequence $(Y_n)$ is uniformly integrable.