The proof should follow as this. And we MUST assume that $X$ and $Z$ are both $L^1$ in order to prove that. A counter example is
$Z_n=3n\circ 1_{\{{0<x\leq\frac{1}{n}}\}}$,$Y_n=2n\circ 1_{\{{0<x\leq\frac{1}{n}}\}}$,$X_n=n\circ 1_{\{{0<x\leq\frac{1}{n}}\}}$
Then $Z_n\rightarrow0$ and $Y_n\rightarrow0$ and $X_n\rightarrow0$ poitnwisely. However $EY_n=2\neq 0$. In fact many counterexamples can be found as long as the sequence of measures corresponding to $Z_n$ are not tight.
Proof:$Z_n-Y_n\geq0$ is a sequence of nonnegative random variables by assumption $Y_n\leq Z_n$. And the sequence is pointwisely convergent to $Z-Y$ by assumption. Fatou's lemma said that $$E(\lim \inf Z_n-Y_n)\leq \liminf E(Z_n-Y_n)=\liminf E(Z_n-Y_n)=\liminf E(Z_n)-\liminf E(Y_n)=EZ-\liminf E(Y_n)$$ (1)
But at the same time $$E(\lim inf Z_n-Y_n)=E(\lim Z_n-Y_n)=E(Z-Y)=EZ-EY$$ (2)
By (1) and (2) we have $$EZ-EY\leq EZ-\lim E(Y_n)$$(3) and since $EZ<\infty$ we have $EY\geq \liminf E(Y_n)$. Apply Fatou's Lemma again $$EY=E(\liminf Y_n)\leq \liminf E(Y_n)$$(4)
(3) and (4) result in $EY=\liminf E(Y_n)$
Symmetric argument involving $X_n$ will lead to $EY=\limsup E(Y_n)$ so $\lim E(Y_n)=EY$ is proved. $\square$
This lemma can actually be used to prove Dominated convergence theorem.
Another (possibly) easier proof is that by noticing $L^1$-convergence is equivalent to $L^1$-Cauchy sequence. Pratt's lemma reveals the essence of "dominance", the dominance family passes the bound on $L^1$ norm to the $X_n$ via Fatou's lemma.