This comes from Tao's notes on probability theory: https://terrytao.wordpress.com/2015/10/03/275a-notes-1-integration-and-expectation/comment-page-1/#comment-681077
Theorem $25$ (Convergence for random variables with bounded moment) Let ${X_1,X_2,\dots}$ be scalar random variables which converge almost surely to a limit ${X}$. Suppose there is ${\varepsilon>0}$ and ${M>0}$ such that ${{\bf E} |X_n|^{1+\varepsilon} \leq M}$ for all ${n}$. Then
$\displaystyle \lim_{n \rightarrow \infty} {\bf E} X_n = {\bf E} X$.
Proof: By taking real and imaginary parts we may assume that the ${X_n}$ (and hence ${X}$) are real-valued. For any natural number {m}, let ${X_n^{[m]}}$ denote the truncation ${X_n^{[m]} := \max(\min(X_n,m),-m)}$ of ${X_n}$ to the interval ${[-m,m]}$, and similarly define ${X^{[m]} := \max(\min(X,m),-m)}$. Then ${X_n^{[m]}}$ converges pointwise to ${X^{[m]}}$, and hence by the bounded convergence theorem
$\displaystyle \lim_{n \rightarrow \infty} {\bf E} X_n^{[m]} = {\bf E} X^{[m]}$.
On the other hand, we have
$\displaystyle |X_n - X_n^{[m]}| \leq m^{-\varepsilon} |X_n|^{1+\varepsilon}$
and thus on taking expectations and using the triangle inequality
$\displaystyle {\bf E} X_n = {\bf E} X_n^{[m]} + O( m^{-\varepsilon} M )$
where we are using the asymptotic notation ${O(X)}$ to denote a quantity bounded in magnitude by ${CX}$ for an absolute constant ${C}$. Also, from Fatou’s lemma we have
$\displaystyle {\bf E} |X|^{1+\varepsilon} \leq M$
so we similarly have
$\displaystyle {\bf E} X = {\bf E} X^{[m]} + O( m^{-\varepsilon} M )$
Putting all this together, we see that
$\displaystyle \liminf_{n \rightarrow \infty} {\bf E} X_n, \limsup_{n \rightarrow \infty} {\bf E} X_n = {\bf E} X + O( m^{-\varepsilon} M )$.
Sending ${m \rightarrow \infty}$, we obtain the claim. $\Box$
My question regards the extra $O( m^{-\varepsilon} M )$ term on the RHS of the last display. Given what has been established above, shouldn't one be able to deduce directly that $\displaystyle \liminf_{n \rightarrow \infty} {\bf E} X_n = \liminf_{n \rightarrow \infty} {\bf E} X_n^{[m]} + O( m^{-\varepsilon} M ) = {\bf E} X^{[m]} + O( m^{-\varepsilon} M ) = {\bf E} X$?