Let $(f_{n})$ a sequence of measurable functions, which converge to $f$ almost everywhere and for which there is a integrable function $g$, such that $|f_{n}|\leq g$ for all $n$. To show that $$\lim\limits_{n\rightarrow \infty}\int_{\Omega}f_{n}=\int_{\Omega}f$$ by Vitali Convergence Theorem we need to show that $(f_{n})$ is uniformly integrable and tight.
In fact, just show that $g$ is uniformly integrable. Given $\epsilon>0$, since $g$ is integrable non-negative, there is $g_{\epsilon}$ measurable bounded function of finite support for which $0\leq g_{\epsilon}\leq g$ and $$0\leq \int_{\Omega}g-\int_{\Omega}g_{\epsilon}< \frac{\epsilon}{2}.$$ Now, let $A\subset \Omega$ measurable set. Since $g-g_{\epsilon}\geq 0$, we have $$\int_{A}g-\int_{A}g_{\epsilon}=\int_{A}[g-g_{\epsilon}]\leq \int_{\Omega}[g-g_{\epsilon}]=\int_{\Omega}g-\int_{\Omega}g_{\epsilon}< \frac{\epsilon}{2}$$ Furthermore, since $g_{\epsilon}$ is bounded, there is $M>0$ such that $0\leq g_{\epsilon}\leq M$, so $$\int_{A}g<\int_{A}g_{\epsilon}+\frac{\epsilon}{2}\leq \int_{A}M+\frac{\epsilon}{2}=M\cdot m(A)+\frac{\epsilon}{2}$$ Thus, taking $\delta=\frac{\epsilon}{2M}$, we have $$m(A)<\delta \Rightarrow \int_{A}g<M\cdot \delta +\frac{\epsilon}{2}=\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$ Therefore, $g$ is uniformly integrable, hence $(f_{n})$ is also, because $$\int_{A}|f_{n}|\leq \int_{A}g<\epsilon.$$ It remains to be shown that $(f_{n})$ is tight. To this end, we will show that $g$ is tight. But its very simple, because given $\epsilon>0$, there is $g_{\epsilon}$ as before such that $0\leq g_{\epsilon}\leq g$ and $$\int_{\Omega}g-\int_{\Omega}g_{\epsilon}<\epsilon$$ and $g_{\epsilon}$ vanishes outside a subset $A_{0}\subset \Omega$ of finite measure. Hence, $$\int_{\Omega-A_{0}}g=\int_{\Omega-A_{0}}[g-g_{\epsilon}]\leq \int_{\Omega}[g-g_{\epsilon}]=\int_{\Omega}g-\int_{\Omega}g_{\epsilon}<\epsilon$$ Therefore, $g$ is tight and so $(f_{n})$ is tight, because $$\int_{\Omega-A_{0}}|f_{n}|\leq \int_{\Omega-A_{0}}g<\epsilon.$$ Hence, we can apply the Vitali Convergence Theorem, concluding that $f$ is integrable and $$\lim\limits_{n\rightarrow \infty}\int_{\Omega}f_{n}=\int_{\Omega}f.$$ This show the Dominated Convergence Theorem.
I would like to know another alternative proofs of Dominated Convergence Theorem.