Riemann Integrable and Uniform Convergence 
Prove that if $f_n \to f_0$ uniformly on $[a,b]$, and all of these functions are Riemann integrable, then $\lim_{n\to \infty} \int_a^b f_n(x) \, dx = \int_a^x f_0(x) \, dx.$

I would try to say since we have uniform convergence we know,
$\lim_{\ n\to\infty}\int_{a}^{b}f_n{(x)}=\int_{a}^{b}f_0{(x)}\ dx$ 
$\Rightarrow \lim_{\ n\to\infty}\int_{a}^{b}f_n{(x)}-\int_{a}^{b}f_0{(x)}\ dx=0$
$\Rightarrow \lim_{\ n\to\infty}\int_{a}^{b}f_n{(x)}-f_0{(x)}\ dx=0$
$\Rightarrow \int_{a}^{b}[\lim_{\ n\to\infty}(f_n(x)-f_0(x))]=0$ ($\textbf{unif convg})$.
Morso, $\sup|(f_n-f_0)|<\epsilon$
Here, the integrand is "$\textbf{0}$" since by assumption $f_n\to f_0$.
I'm not sure about this.
 A: Let $\epsilon>0$ and choose $N$ such that if $n \ge N$, then $|f(x)-f_n(x)| \le { \epsilon \over b-a}$ for all $x \in [a,b]$.
Then $\int_a^b |f(x)-f_n(x)| dx \le \int_a^b { \epsilon \over b-a} dx = \epsilon$.
Hence $\lim_{n \to \infty} \int_a^b |f(x)-f_n(x)| dx = 0$.
Since $|\int_a^b (f(x)-f_n(x))dx | \le \int_a^b |f(x)-f_n(x)| dx$, and
$\int_a^b (f(x)-f_n(x))dx = \int_a^b f(x)dx - \int_a^b f_n(x) dx$, we see that 
$\lim_{n \to \infty}\int_a^b  f_n(x) dx = \int_a^b f(x)dx$.
A: I think it fits here: If $(f_n)$ is a sequence of Riemann-integrable functions on $[a,b]$ with $f_n \rightrightarrows f$, then $f$ is also Riemann-integrable on $[a,b]$.
Let $\varepsilon > 0$ be given. Then from the uniform convergence, we have that (for big enough $n$)
$$-\frac{\varepsilon}{3(b-a)} +f_n< f < \frac{\varepsilon}{3(b-a)}+f_n$$
Which means that
$$\underline{\int}\left(-\frac{\varepsilon}{3(b-a)} +f_n\right)\leqslant \underline{\int} f \leqslant \overline{\int} f \leqslant \overline{\int} \left(\frac{\varepsilon}{3(b-a)}+f_n\right)$$
Which means that
$$\int f_n -\frac{\varepsilon}{3} \leqslant \underline{\int} f \leqslant \overline{\int} f \leqslant \int f_n + \frac{\varepsilon}{3}$$
I.e.
$$\left|\overline{\int} f -\underline{\int} f\right| \leqslant \left|\overline{\int}f -\int f_n\right|+\left|\underline{\int} f-\int f_n \right|\leqslant \frac{\varepsilon}{3}+\frac{\varepsilon}{3} < \varepsilon$$
Which means that
$$\underline{\int} f = \overline{\int} f$$
So $f$ is Riemann-integrable.
