sequences and series - Lebesgue integration and uniform convergence

Let $\Omega$ be a bounded and measurable set in $\mathbb{R}$.

If $\{f_n\}$ is a sequence of bounded and Lebesgue integrable functions on $\Omega$. If $f_n$ uniformly converges to $f$, then how to proof that

1) $f$ is Lebesgue integrable on $\Omega$.

2) $\int_{\Omega}f_n d\mu\rightarrow\int_{\Omega}f d\mu$
, i.e
$$
\lim\limits_{n\rightarrow\infty}\int_{\Omega}f_n d\mu=\int_{\Omega}\lim\limits_{n\rightarrow\infty}f_n d\mu=\int_{\Omega}f d\mu
$$

Answer

The limit of a sequence of measurable functions is measurable. Choose $N$ with the property that $|f_N(x) - f(x)| < 1$ for all $x \in \Omega$. Then $|f(x)| < |f_N(x)| + 1$ so that
$$
\int_\Omega |f| \, d\mu \le \int_\Omega |f_N| \, d\mu + \mu(\Omega) < \infty$$ implying $f$ is integrable. Moreover, for any $n$ you have
$$ \left| \int_\Omega f \, d\mu - \int_\Omega f_n \, d\mu \right| \le \int_\Omega |f - f_n| \, d\mu \le \left( \sup_{x \in \Omega} |f(x) - f_n(x)| \right)\mu(\Omega)$$

and the right hand side converges to $0$ by the definition of uniform convergence. Thus $$\int_\Omega f_n \, d\mu \to \int_\Omega f \, d\mu.$$