So the question is, for Lebesgue probability space $(\mathbb{R}, B(\mathbb{R}), \lambda)$, define $$X_n(\omega) = \begin{cases}n^2\omega, &0\leq \omega \leq \frac{1}{n}, \\ 2n-n^2\omega, &\frac{1}{n} \leq \omega \leq \frac{2}{n}, \\ 0, &\frac{2}{n} \leq \omega \leq 1 \end{cases}$$ I am tasked to show that $X_n$ is point-wise convergent to $0$, as $n$ tends to $\infty$, while $EX_n \not\rightarrow EX$.
So for the first task my first thought was to using the definition of pointwise convergence, i.e., $\forall \ \epsilon>0,$ there exists an index $N$ such that $\forall n>N, \ \ |X_n-X|< \epsilon$ at each critical point, i.e., $ \omega_0=1/n, \ \omega_1=2/n $ as each function defined in each boundary is linear function therefore continuous.
However, considering in the case of $X$, where $n$ tended to $\infty$, as you can see I cannot write the function because $n$ exploded, and since the fact that $\omega_0, \omega_1$ are dependent on $n$ may prevent me to proceed this way? or because $n$ already tended to $\infty$ I can just consider $X=0$ $\forall \omega$?
Also any hint to proceed with the first and the second part would be greatly appreciated. I think $EX$ would be evaluated to $0$. so I need to show that $EX_n \not\rightarrow 0$ at the essence, but I am not sure how am I gonna do that.
Thank you.
To determine pointwise convergence you have to examine the sequence $(X_n(\omega))$ for each fixed $\omega$ and observe what happens as $n\to\infty$. That is, you fix an $\omega\in[0,1]$ and study the sequence of real numbers $(X_n(\omega))$. To do so you have to observe how the functions $X_n$ change as $n$ increases. In particular, you'll see that if your fixed $\omega$ is greater than zero, then for all large enough $n$ the region $[0,\frac2n]$ where $X_n$ is nonzero eventually slides to the left of $\omega$. In other words, for all large enough $n$, $X_n(\omega)=0$. The case $\omega=0$ is handled separately. In either case you will have shown that $X_n(\omega)\to0$ as $n\to\infty$.
As for your second part, you need to compute $E(X_n)$, which is the area under $X_n$, for each $n$. This is a straightforward exercise in integration. As for $E(X)$, this is zero since $X=0$.
It will help to draw a picture of $X_n$. Observe that the graph of $X_n$ as a function of $\omega$ is a tent with base $[0,\frac2n]$ and peak of height $n$ at $\omega=\frac1n$, and zero elsewhere. (I'm assuming in the second case you have a typo and the formula should be $2n-n^2\omega$.)