If $f_n$ converges uniformly to $f$, then $\int f_n\ d\mu \to \int f\ d\mu$ as $ n \to \infty $ Let ($\Omega, \mathcal{F}, \mu)$, with $\mu (\Omega) <\infty$ and suppose  $f_n , n \geq 1$ is a sequence of integrable functions that converges uniformly on $\Omega$ to $f$. Given that $f$ is integrable, show that  $$\int f_n\ d\mu  \to  \int f\ d\mu \quad\text{ as }\quad n \to \infty $$
I tried :
$\left|\int_\Omega f_n \, d\mu - \int_\Omega f\, d\mu\right| $
: (By linearity of the integral)
$= |\int_\Omega (f_n - f)\, d\mu|\leq \int_\Omega |f_n - f|\, d\mu$
How do we prove the inequality above?
I tried the following but  I am not sure if it's correct or sufficient mathematical proof.
Since $f_n$ integrable, so  $\int_\Omega f_n \ d\mu$ is finite which means it exists.
$f$ integrable, so$\int_\Omega f\ d\mu$ is finite,   which means it also exists.
Thus, since they both exist  $(f_n - f)$ exists   $= |\int_\Omega (f_n - f)\, d\mu|$ is finite.
Hence from basic property of integral since  $= |\int_\Omega (f_n - f)\, d\mu|$ exists
Finally  how do we apply the Uniform convergence on $  \int_\Omega |f_n - f|\, d\mu$ to conclude that
$\int_\Omega f\ d\mu  \to  \int_\Omega f\ d\mu$ as $ n \to \infty $ ?
 A: Since $f_n$ converges uniformly on $\Omega$ to $f$, for $\forall \varepsilon>0$, there is $N>0$ which is independent of $x\in\Omega$ such that
$$ |f_n(x)-f(x)<\frac{\varepsilon}{\mu(\Omega)}, \forall n\ge N, x\in\Omega. $$
So for above $\varepsilon>0$ and $N>0$, when $n\ge N$,
$$\bigg|\int_\Omega (f_n - f)\, d\mu\bigg|\le\int_\Omega |f_n-f|\ d\mu\le\int_{\Omega}\frac{\varepsilon}{\mu(\Omega)}d\mu=\varepsilon $$
which implies
$$ \lim_{ n \to \infty}\int_\Omega f_n \, d\mu=\int_\Omega f_n \, d\mu. $$
A: Step 1.
$$
\left|\,\int_\Omega f\,d\mu \,\right|\le \int_\Omega |f|\,d\mu
$$
Proof. Set
$$
f_-=\max\{-f,0\}, \quad
f_+=\max\{f,0\}.
$$
Then $f_-,f_+$ are measurable, $f_-,f_+\ge 0$ and also
$f=f_+-f_-$ and $|f|=f_++f_-$. So
$$
\left|\,\int_\Omega f\,d\mu \,\right|=\left|\,\int_\Omega (f_--f_-)\,d\mu \,\right|
=\left|\,\int_\Omega f_+\,d\mu -\int_\Omega f_-\,d\mu \,\right| \le
\left|\,\int_\Omega f_+\,d\mu\,\right| +\left|\,\int_\Omega f_-\,d\mu \,\right|
\\=\int_\Omega f_+\,d\mu+\int_\Omega f_-\,d\mu =\int_\Omega (f_-+f_-)\,d\mu
=\int_\Omega |f|\,d\mu.
$$
Step 2. If $f_n\to f$ uniformly on $\Omega$, then for every $\varepsilon>0$, there exists a $N$, such that
$$
n\ge N \quad\Longrightarrow\quad |f_n(x)-f(x)|<\frac{\varepsilon}{\mu(\Omega)+1}
$$
for all $x\in\Omega$.
Hence, if $n\ge N$, then
$$
\left|\,\int_\Omega f_n\,d\mu-\int_\Omega f\,d\mu\,\right|=
\left|\,\int_\Omega (f_n-f)\,d\mu\,\right|\le
\int_\Omega |f_n-f|\,d\mu \le \int_\Omega \frac{\varepsilon}{\mu(\Omega)+1}\,d\mu=\frac{\varepsilon\, \mu(\Omega)}{\mu(\Omega)+1}<\varepsilon.
$$
and thus
$$
\lim_{n\to\infty}\int_\Omega f_n\,d\mu=\int_\Omega f\,d\mu.
$$
