Question:
Let $X_n$, n = 1,... denote a sequence of real-valued random variables; $X_n$ is said to converge in mean if
$\hspace{20mm}$$$\lim_{n\to\infty} E[|X_n-X|] = 0$$
Show that if $X_n$ converges to X in mean and $E[|X|] < \infty$, then
$\hspace{20mm}$ $$\lim_{n\to\infty} E[X_n] = E[X]$$
I'm not entirely sure about this one, but I think the triangle inequality can be used:
$\hspace{20mm}$$$\lim_{n\to\infty}E[|X_n-X|] \geq \lim_{n\to\infty}E[|X_n|]-E[|X|] = 0$$
Use the following integral property :
$$\big\vert \mathrm{E}\left[ Y \right] \big\vert \leq \mathrm{E}\left[ \vert Y \vert \right]$$
By taking ($Y=X_{n}-X$) and using the linearity of the expectation, we get :
$$\big\vert \mathrm{E}\left[ X_{n} \right] - \mathrm{E}\left[ X \right] \big\vert = \big\vert \mathrm{E}\left[ X_{n}-X \right] \big\vert \leq \mathrm{E}\left[ \big\vert X_{n} - X \big\vert \right]$$
Since we know that $\displaystyle \lim \limits_{n \to +\infty} \mathrm{E}\left[ \big\vert X_{n}-X \big\vert \right] = 0$, we get :
$$ \lim \limits_{n \to +\infty} \mathrm{E}\left[X_{n}\right] = \mathrm{E}\left[X\right] $$