I have two questions on mode of convergence: Let us consider the following sequence of r.v.s $X_n=\frac{n+2}{n+1}\omega$ for $n\geq 0$. Clearly $X_n$ converges a.s. to $X=\omega$ and also in $L^1$. But does $X_n$ converges in $L^2$ to$X=\omega$? I think so, indeed$$E[|X_n-X|^2]=(\frac{n+2}{n+1})^2+1-2\frac{n+2}{n+1}]E[\omega^2]$$ which clearly goes to $0$ as $n$ goes to $\infty$. Is this correct?
Does convergence in $L^2$ implies convergence in $L^1$? Also if $X_n$ does not converge in $L^1$ then it will not converge in $L^2$ for sure. Am I right?
$\begin{align} E[|X_n - X|^2] &= E[|(\frac{n+2}{n+1}) \omega - \omega|^2] \\ &= E[\frac{\omega^2}{(n+1)^2}]\\ &= \frac{1}{(n+1)^2} E[\omega^2] \end{align}$.
Clearly, this goes to zero as $n \rightarrow \infty$. Hence, $X_n$ converges to $ X$ in $L^1$.
Convergence in $L^2 $ implies convergence in $L^1$ follows from Holder's inequality.