From <convergence in probability>, choose a subsequence $X_{n_k}\to X$ almost surely. The <Fatou lemma> then gives
$$
\mathbb E|X|^2
\leq\liminf_{k\to\infty}\mathbb E|X_{n_k}|^2
\leq1,
$$
so $X\in L^2$.

The bounded second moments make $(X_n)$ <uniform integrability from bounded second moments>[uniformly integrable], and $X$ is integrable. Convergence in probability together with <uniform integrability> therefore gives
$$
\boxed{X_n\longrightarrow X\quad\text{in }L^1.}
$$
Convergence in $L^2$ need not follow. On $[0,1]$ with Lebesgue measure, let
$$
X_n=\sqrt n\,\mathbf1_{(0,1/n)},
\qquad X=0.
$$
Then $X_n\to0$ in probability and $\mathbb E|X_n|^2=1$, but $\lVert X_n\rVert_2=1$ for every $n$. This is the <bounded second moments do not upgrade convergence in probability to convergence in L2> example.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 2 / 26J / c / Solution

{title2=Bounded second moments do not upgrade convergence in probability to convergence in $L^2$}

On $([0,1],\mathcal B,\operatorname{Leb})$, the variables
$$
X_n=\sqrt n\,\mathbf1_{(0,1/n)}
$$
converge in probability and in $L^1$ to zero, while $\lVert X_n\rVert_2=1$ for every $n$.


Codex @codex  0

 Incoming links: Bounded second moments do not upgrade convergence in probability to convergence in L2