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 \begin{equation}\mathbb{P}(\omega \in \Omega: \lim_{n \rightarrow \infty}X_n = X) = 1 \Leftrightarrow X_n \rightarrow X\text{ a.s.}\end{equation}
NTS: \begin{equation} \lim_{n \rightarrow \infty}\mathbb{E}|X_n| \rightarrow E|X|<\infty \Rightarrow \lim_{n\rightarrow \infty}\mathbb{E}[|X_n - X|^1 ] =0.\end{equation} 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?