I want to prove the following:
$$ \lim_{n \rightarrow \infty} \int_{0}^{b_{n}} f_{n}(x) dx= \int_{0}^{1} f(x) dx $$ provided that $f_{n}$ is Riemann-integrable on $[0,1]$ and $f_{n} \rightarrow f$ uniformly on $[0,1]$ and $b_{1} \leq b_{2}...\leq b_{n}\le \dots$ and $b_{n} \rightarrow 1$.
I started the proof by saying that since $f_{n} \rightarrow f$ uniformly so we can exchange the limit and the integral. But, that seems to be making it almost trivial. Am I missing something very important?
Use $$\left|\int_0^{b_n}f_n-\int_{0}^{1}f\right|\leq \left|\int_0^{b_n}f_n-\int_{0}^{1}f_n\right|+\left|\int_0^{1}f_n-\int_{0}^{1}f\right|$$