Convergence of sequence of random variables in $L_1$ norm

Question

I am struggling with the following problem: Given is the probability space $(\Omega, \mathcal{F}, \mu)$ with $\Omega = [0,\infty), \mathcal{F} = \mathcal{M}_{[0,\infty)}$ and measure $\mu$ given by

\begin{equation} \mu([a,b]) = \int\limits_a^b \dfrac{1}{(x+1)^2}dm \end{equation}

and the sequence of random variables

\begin{equation} X_n(\omega) = \max{(0,n^2 - n^5 |\omega - n|)}. \end{equation}

I am trying to determine if $X_n \rightarrow 0$ in $L_1$ norm. To this end, I tried first to find the distribution function $F_{X_n}$ of $X_n$. I obtained:

\begin{equation} F_{X_n}(y) = \mu ( (-\infty, \dfrac{y-n}{n^5} + n) \cup (\dfrac{n^2 - y}{n^5} + n, \infty))\\ = \dfrac{n^5}{n^2 - y + n^6 + n^5} - \dfrac{n^5}{y-n^2+n^6+n^5} \end{equation}

if $y < 0$ and 1 otherwise.

Differentiating this yields

\begin{equation} f_{X_n}(y) = \dfrac{n^5}{(n^2 -y + n^6 + n^5)^2} - \dfrac{n^5}{(y-n^2+n^6+n^5)^2} \end{equation}

if $y<0$.

But now

\begin{equation} \int\limits_{-\infty}^{\infty} f_{X_n}(y) dy \neq 1. \end{equation}

Have I just made a mistake? Or is my entire approach incorrect? Could there be an easier solution? Thanks in advance!

Answer 1

Hint: $X_n(\omega) = 0$ outside the interval $(n - 1/n^3, n+1/n^3)$. This lead to a very simple upper bound for $\|X_n\|_1$