A variant of the dominated convergence theorem in probability theory The problem is from Durrett's book. Let $X_n\to X$ a.s. and let $g:\mathbb R\to (0,\infty)$ be a continuous function so that $|x|/g(x)\to 0$ as $|x|\to \infty$. We also assume $Eg(X_n)\leq C<\infty$. Show that $EX_n\to EX$.
My attempt is to use Egorov's theorem. The set (denoted by $A$) on which the convergence is uniform is easy to handle. To handle the other part, I would like to estimate it as follows:
$$
\int_{A^c}|X_n-X|dP\leq \int_{A^c}\frac {|X_n-X|}{g(X_n-X)}g(X_n-X)dP.
$$
If $|X_n-X|$ is small, then it is easy to do. If $|X_n-X|$ is large, we use the assumption that $|x|/g(x)\to 0$. This gives us the only difficulty, which is to show that $Eg(X_n-X)<\infty$. But how to do that?
Edited: It would be done if we assume in addition that $g(x)$ is increasing as $x\to \infty$, which is usually the case in applications. However, the continuity of $g$ is not used in this case. 
 A: I was thinking along the lines of "uniform integrability" rather than "Egorov theorem."  Using $g(x)=xm(x)$ for $x \geq 1$ from  my comment above, we get for all $n$ and all $r\geq 1$:
$$ C \geq E[g(X_n)] \geq E[g(X_n)|X_n\geq r]P[X_n\geq r] = E[X_nm(X_n)|X_n\geq r]P[X_n\geq r] \geq ...$$
or you can modify the argument to treat $|X_n|\geq r$ (the above just bounds one side if $X_n$ is real-valued rather than positive-valued).  
I was also focusing more on the non-decreasing property of $\inf_{y \geq x} m(y)$ over $x \geq 1$, rather than the increasing property of $x \inf_{y \geq x} m(y)$ over $x \geq 1$. 
A: Thanks to the idea of Michael, my idea is to use the Vitali convergence theorem, hence it suffices to show that $\{X_n\}$ is uniformly integrable.
Let $\epsilon>0$, then there is $A>0$ so that if $a\geq A$ and $|X_n|\geq a$, $|X_n|/g(X_n)<\epsilon$. Then for $a\geq A$, we have for any $n$,
        $$
   E|X_n|1_{(|X_n|\geq a)}=E\frac {|X_n|}{g(X_n)}g(X_n)1_{(|X_n|\geq a)}\leq \epsilon Eg(X_n)1_{(|X_n|\geq a)}\leq \epsilon C.
   $$
A: The problem can be transformed to "a.s convergence + condition on domination implies L1 convergence" by considering subsequences.
Fact 1:

Given a sequence $y_n$ on some topological space. If every subsequence of $y_{n(m)}$ has a further subsequence $y_{n(m_k)}$ that converges to $y$ then $y_n \rightarrow y$. (Durrett 5th ed, Thm 2.3.3).

So it suffices to consider subsequence $X_{n(m)}$ and show it has a further subsequence that converges in L1.
Fact 2:

$X_n \rightarrow X$ in probability iff every subsequence $X_{n(m)}$ has a further subsequence $X_{n(m_k)}$ that converges almost surely to $X$. (Durrett 5th ed, Thm 2.3.2).

Take $X_{n(m_k)} \rightarrow X$ a.s. Fix $\epsilon > 0$, then there exits $\delta >0 $ such that $|x|>\delta \implies |x|/g(x) < \epsilon$.
Now $X_{n(m_k)} = X_{n(m_k)} \mathbb{1}_{|X_{n(m_k)}|>\delta} + X_{n(m_k)} \mathbb{1}_{|X_{n(m_k)}|\leq \delta}$.
By Dominated Convergence Theorem, $EX_{n(m_k)} \mathbb{1}_{|X_{n(m_k)}|\leq \delta} \rightarrow E|X|\mathbb{1}_{|X|\leq \delta}$.
Also $|E X_{n(m_k)} \mathbb{1}_{|X_{n(m_k)}|>\delta}|\leq E|X_{n(m_k)}|\mathbb{1}_{|X_{n(m_k)}|>\delta} \leq \epsilon Eg(X_{n(m_k)}) \leq \epsilon C$.
We have $$\limsup |EX_{n(m_k)} - EX| \leq \limsup \{|EX_{n(m_k)} \mathbb{1}_{|X_{n(m_k)}|\leq \delta} - EX \mathbb{1}_{|X|\leq \delta}| + |E X \mathbb{1}_{|X|>\delta}|+|E X_{n(m_k)} \mathbb{1}_{|X_{n(m_k)}|>\delta}|\}$$
$$ \leq |EX\mathbb{1}_{|X|>\delta}| + \epsilon C$$
To show $|EX\mathbb{1}_{|X|>\delta}|$ vanish as $\delta \rightarrow \infty$, we have $$E(g(X)) = E(\liminf g(X_{n(m_k)})) \leq \liminf E(g(X_{n(m_k)})) \leq C < \infty,$$ using continuity of g to get the first equality. We have $P(|X| = \infty) \leq P(g(X) = \infty) = 0$ since $g(X)$ is L1.
So as $\epsilon \rightarrow 0$ and $\delta \rightarrow \infty$, $\limsup |EX_{n(m_k)} - EX| = 0$.
Then $EX_n \rightarrow EX$ from Fact 2.
