Let the sample space be $[0,1]$ with the Borel sigma algebra and the probability dx:
$X(\omega)= \frac {1} {1-\omega}$.
Define $X_n(w) = min(X(w),n)$. Show that $X_n \uparrow X$ $a.s.$, and that $E[X_n]\uparrow E[X]$
Here is my attempt. First, I know that $E[X]=+\infty$. From the definition of $X_n$ we can see that :
$$ X_n(w) = \begin{cases} X(w), & \text{for $ 0\le w \le \frac {n-1}{n}$ } \\ n, & \text{for $\frac {n-1}{n}< w \le 1$ } \\ \end{cases} $$
From this we can see that $X_n \uparrow X$ $a.s.$. Then I can use the monotone convergence theorem to claim that $E[X_n]\uparrow E[X]$. But I want to really check that $E[X_n]\uparrow E[X]$ by computing $E[X_n]$. Is the following expression correct? $$E[X_n]=\int_0^{\frac {n-1}{n}} \frac 1 {1-\omega} d\omega + \int_\frac {n-1}{n}^{1} n d\omega$$ Is this correct? How do I show that this sequence converges to $E[X]$? I think I should take the limit, but it looks like it does not work.
What you have done is correct. To see that $EX_n \to \infty$ just note that (by direct evaluation of the integral) $EX_n \geq \int_0^{1-\frac 1 n} \frac 1 {1-\omega} d\omega =\ln n \to \infty$.
$\int_0^{1-\frac 1 n} \frac 1 {1-\omega} d\omega =-\ln (1-\omega)|_0^{1-\frac 1 n} =-\ln (\frac 1n)=ln \, n$