modified Dominated convergence theorem Dominated convergence theorem:
If for some random variable $Z$, we have that $|X_{n}|\leq |Z|$ for all $n$ and $\mathbb{E}[Z]<\infty$, then $X_{n}\stackrel{d}{\to}X$ implies that $\mathbb{E}[X_{n}] \to \mathbb{E}[X]$. This is well known.
I got stuck with the following "modification": assume, that instead of $|X_{n}|\leq |Z|$ we have the following:
$|X_{n}|\leq |Z_{n}|$, where $Z_{n}$ is a sequence s. t. $Z_{n}\stackrel{d}{\to}Z$, $\mathbb{E}[Z_{n}]\to \mathbb{E}[Z]$ and $\mathbb{E}[Z_{n}]=a<\infty$, THE SAME! for all $n$. 
Is the statement about convergence of means true in this case?
I am thinking to introduce a new variable $Z_{max} = \max\{Z_{1}, Z_{2}, \dots, Z_{n}, \dots\}$, but I doubt that $\mathbb{E}[Z_{max}]<\infty$ in this case... 
 A: Given the counter-example in my previous answer, let's modify the statement:  
Claim 1: (Modified statement)
Suppose $X_n\rightarrow X$ in distribution, $Z_n\rightarrow Z$ in distribution, $E[Z_n]\rightarrow E[Z]<\infty$, and $|X_n|\leq Z_n$ for all $n \in \{1, 2, 3, ...\}$. Then $E[X]$ is finite and $E[X_n]\rightarrow E[X]$. 
Before proving this, it is useful to state and prove a lesser-known modification of the usual Lebesgue dominated convergence theorem: 
Claim 2: (Modified LDC)
Suppose $f_n$ and $g_n$ are real-valued functions that satisfy $0\leq f_n(x)\leq g_n(x)$ for all $x \in \mathbb{R}$. Suppose that $f_n$ converges pointwise almost everywhere to a function $f$, $g_n$ converges pointwise almost everywhere to a function $g$, and $\int  g_n  \rightarrow \int g  < \infty$. Then $\int  f  < \infty$ and $\int  f_n  \rightarrow \int f$. 
Proof of Claim 2: (almost the same as the standard LDC proof) We know by Fatou's lemma: 
$$ \int \liminf f_n \leq \liminf \int f_n  $$ 
The left-hand-side is $\int f$.  Hence: 
$$ \int f \leq \liminf \int f_n \quad (Eq 1) $$
It also easily follows that $\int f < \infty$. 
On the other hand, applying Fatou to the nonnegative functions $g_n-f_n$ gives: 
$$  \underbrace{\int \liminf (g_n-f_n)}_{\int g - \int f } \leq \underbrace{
\liminf\int (g_n-f_n)}_{\int g - \limsup \int f_n} $$
Since $\int g$ is finite we get 
$$\limsup \int f_n \leq \int f $$
Combining this with equation (1) gives: 
$$ \int f \leq \liminf \int f_n \leq \limsup \int f_n \leq \int f $$ 
$\Box$
Derivation of Claim 1:
It suffices to assume $X_n\geq 0$ for all $n$ (else, we can write $X_n=X_n^+ - X_n^-$ and repeat the argument on the positive and negative parts). Define: 
\begin{align}
f_n(x) &= P[X_n>x]\\
f(x) &= P[X>x]\\
g_n(x) &= P[Z_n>x]\\
g(x) &=P[Z>x]
\end{align}
We know that $0\leq f_n(x) \leq g_n(x)$ for all $n \in \{1, 2, 3, ...\}$ and all $x \in \mathbb{R}$. We also know that $f_n$ converges to $f$ pointwise almost everywhere, and $g_n$ converges to $g$ pointwise almost everywhere. We also know that:
\begin{align}
\int_0^{\infty} g_n(x)dx = E[Z_n]\\
\int_0^{\infty} f_n(x)dx = E[X_n]\\
\int_0^{\infty} g(x)dx = E[Z] < \infty \\
\int_0^{\infty} f(x)dx = E[X]
\end{align}
Finally, we know that $\int_0^{\infty} g_n(x)dx\rightarrow \int_0^{\infty} g(x)dx$. It follows by Claim 2 that $\int_0^{\infty} f(x)dx< \infty$ and $\int_0^{\infty} f_n(x)dx \rightarrow \int_0^{\infty} f(x)dx$.  That is, $E[X_n]\rightarrow E[X]$. $\Box$
A: I believe this is true under a minor modification, but it is not true as stated: For $n \in \{1, 2, 3, …\}$ define $Z_n$ by: 
$$ Z_n = \left\{ \begin{array}{ll}
n &\mbox{ with prob $1/(2n)$} \\
-n &\mbox{with prob $1/(2n)$} \\
0  & \mbox{ otherwise} 
\end{array}
\right.$$
Then $Z_n\rightarrow 0$ in distribution and $E[Z_n]=0$ for all $n$.  Define $X_n=|Z_n|$ for all $n$.  Then $X_n\rightarrow 0$ in distribution, but $E[X_n]=1$ for all $n$.  
A: I propose the following solution:
Let $Z_{n}>0$.
Since $Z_{n}$ is a sequence s. t. $Z_{n}\stackrel{d}{\to}Z$, $\mathbb{E}[Z_{n}]\to \mathbb{E}[Z]$ and $\mathbb{E}[Z_{n}]=a<\infty$, then $Z_{n}$ is asymptotically uniformly integrable. 
Next, we can show, that, since $X_{n}\leq Z_{n}$, and, let's say that $X_{n}\geq 0$, then $X_{n}$ is also asymptotically uniformly integrable. 
$$
\mathbb{E}[X_{n}\mathbb{I}\{X_{n}>M\}]\leq \mathbb{E}[Z_{n}\mathbb{I}\{X_{n}>M\}]\leq\mathbb{E}[Z_{n}\mathbb{I}\{Z_{n}>M\}],
$$
so that
$$
\limsup \mathbb{E}[X_{n}\mathbb{I}\{X_{n}>M\}] \leq \limsup\mathbb{E}[Z_{n}\mathbb{I}\{Z_{n}>M\}]\to 0,
$$
as $n\to\infty$.
Then, we have done. 
