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.
Best Answer
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 $\lim_{n \to \infty}\int_a^b f_n(x) dx = \int_a^b f(x)dx$.