Does $E(X_n)\rightarrow E(X)$ as $n\to\infty$ If there is a sequence of random variables ${X_n}$ converging almost surely to $X$, therm is it true that $E(X_n)\rightarrow E(X)$ as $n\to\infty$ ? Only thing given is that $E(X_n)\le 23$ for all $n$.
I am not getting how to do it. I can't use DCT here, can I?

No it's not.
Consider: $$X_n(\omega) = \begin{cases}
        n & \mbox{ if } \omega\in [0,\frac{1}{n}] \\
        0  & \mbox{ otherwise}
         \end{cases}$$
Then $X_n \to X$ where $X \equiv 0$, $E[X_n] = 1 \le 23$, so $\lim\limits_{n\to\infty} E[X_n] = 1$ but $E[X] = 0$
