# Almost sure convergence and L1 convergence

I am preparing myself for the mid-term exam of my probability theory exam, and am solving questions from previous years exams. One of these questions I couldn't answer, and so far I haven't found anything similar online.

Suppose that $X_n \rightarrow X$ a.s. as $n \rightarrow \infty$. Prove or disprove that, if $\lim_{n \rightarrow \infty} \mathbb{E}|X_n| \rightarrow \mathbb{E}|X| < \infty$, then $X_n \rightarrow X$ in $L^1$ i.e. $\Bbb E[|X_n-X|] \to 0$.

Here's my current approach: Given $$\mathbb{P}(\omega \in \Omega: \lim_{n \rightarrow \infty}X_n = X) = 1 \Leftrightarrow X_n \rightarrow X\text{ a.s.}$$

NTS: $$\lim_{n \rightarrow \infty}\mathbb{E}|X_n| \rightarrow E|X|<\infty \Rightarrow \lim_{n\rightarrow \infty}\mathbb{E}[|X_n - X|^1 ] =0.$$ Since $X_n \rightarrow X$ almost surely, $X_n \rightarrow X$ in probability. $L^1$ convergence is implied by convergence in probability + uniform integrability, so it suffices to show that $(X_n)$ is uniformly integrable.

To show uniform integrability, I then define a function $f$ that is bounded and continuous, so that $f \circ X_n \rightarrow f \circ X$ in probability, and therefore $\mathbb{E}[f \circ X_n] \rightarrow \mathbb{E}[f \circ X]$, which (together with the assumption that $\mathbb{E}X_n \rightarrow \mathbb{E}X$) implies that $\mathbb{E}[X_n - f \circ X_n] \rightarrow \mathbb{E}[X - f \circ X]$.

Last, fix some $\varepsilon <0$ and use the fact that X is integrable to show that the expectation of X over some interval of the function tends to 0.

For more details, the proof is also given in the book "Probability and Stochastics" by Erhan Cinlar (page 108f, Theorem 4.9).

However, this proof seems a little indirect because I am not "really" using the almost sure convergence, but rather am just working with convergence in probability. Is there any better (i.e more direct) approach?

Lemma: If a sequence of $L^p$ integrable functions $f_n$ converges a.e. to an $L^p$ integrable function $f$ with $p\geq 1$ and $\lim_n \Vert f_n \Vert_p = \Vert f \Vert_p$ holds true, then $\lim_n \Vert f_n - f \Vert_p = 0$.
Proof for p=1 (taken from Kusolitsch (2010)): Consider functions $$f_n^* = \begin{cases} f_n, & \vert f_n \vert \leq \vert f \vert, \\ \vert f \vert sgn(f_n), & \vert f_n \vert > \vert f \vert, \end{cases}$$ which are dominated by the $L^1$ integrable $\vert f \vert$ and converge to $f$ a.e. So the functions $\vert f_n^* - f \vert$ are dominated by $2\vert f \vert$ and vanish a.e., and the dominated convergence theorem yields $$\lim_n \int \vert f_n^* \vert = \int \vert f \vert$$ and also $$\lim_n \int \vert f_n^* - f \vert = 0.$$ By definition $f_n^*$ always has the same sign as $f_n$ and $\vert f_n^* \vert \leq \vert f_n \vert$. So one gets $\vert f_n - f_n^* \vert = \vert f_n \vert - \vert f_n^* \vert$, and $$\int \vert f_n - f_n^* \vert = \int \vert f_n \vert - \int \vert f_n^* \vert.$$ Since both integrals on the right hand side converge to $\int \vert f\vert<\infty$, this yields the conclusion.