# $\underset{n\rightarrow \infty}{\lim}\int_{0}^{b_{n}}f_n(x)dx=\int_0^1f(x)dx$

Let $$(f_n)_{n\in\mathbb{N}}$$ ne a sequence of continuous functions on $$[0,1]\rightarrow\mathbb{R}$$ that converges uniformly to $$f:[0,1]\rightarrow\mathbb{R}$$. Let $$(b_n)_{n\in\mathbb{R}}$$ be an increasing sequence of real numbers in $$(0,1)$$ that converges to 1. Prove that : $$\underset{n\rightarrow \infty}{\lim}\int_{0}^{b_{n}}f_n(x)dx=\int_0^1f(x)dx$$

As far as I know if we have a sequence of functions converging uniformly to $$f$$ AND if they are uniformly bounded we can have the following: $$\underset{n\rightarrow \infty}{\lim}\int_{0}^{1}f_n(x)dx=\int_0^1f(x)dx$$.
But here it is not so. Even the uniform boundedness is also not mentioned.

• Uniform convergence is sufficient for the interchange of limit with the integral. – Math1000 Jan 5 at 22:12
• Thank you for correcting – DD90 Jan 5 at 22:36

For any $$\epsilon > 0$$, we have for all sufficiently large $$n$$ $$\left|\int_0^{b_n} f_n - \int_0^1 f \right| \leqslant \int_0^{b_n}|f_n - f| + \int_{b_n}^1 |f| \leqslant \epsilon b_n + \sup_{x \in [0,1]} |f(x)| (1 - b_n)$$
• Thanks. It works. But I need a small clarification: is it always true that $|\int f dx|\leq\int|f|dx$? If so can you please point out a website or a book which state this result? Thanks again! – DD90 Jan 5 at 22:35
• So $-|f(x)| \leqslant f(x) \leqslant |f(x)|$. One side gives us $\int_0^1f \leqslant \int_0^1|f|$. The other side gives us $-\int_0^1|f| \leqslant \int_0^1 f$. Thus $|\int_0^1 f| \leqslant \int_0^1 |f|$. – RRL Jan 5 at 22:38