# Question about limit of integrals

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?

• You are missing a limit in the second line. Is intergability in the Riemann sense oo Lebesgue sense? Your argument does not take into account the limits of integration. Mar 20 at 23:21
• @KaviRamaMurthy I have fixed the errors now. Mar 20 at 23:30
• Yes. When you "exchange the limit and the integral," the integrals are all occurring on the same interval. Mar 20 at 23:49
• Do you observe that you need to handle the integrand as well as interval of integration as $n$ tends to $\infty$? You can note that $$\int_0^{b_n}f_n(x)\,dx=\int_0^1 f_n(x) \, dx-\int_{b_n} ^{1}f_n(x)\,dx$$ the first term tends to $\int_0^1 f(x) \, dx$ (due to uniform convergence). Can you handle the second term? Mar 21 at 2:33

Use $$\left|\int_0^{b_n}f_n-\int_{0}^{1}f\right|\leq \left|\int_0^{b_n}f_n-\int_{0}^{1}f_n\right|+\left|\int_0^{1}f_n-\int_{0}^{1}f\right|$$