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

Question

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.

Answer 1

Hint:

For any $\epsilon > 0$, we have for all sufficiently large $n$ $$\left|\int_0^{b_n} f_n - \int_0^1 f \right| \leqslant \int_0^{b_n}|f_n - f| + \int_{b_n}^1 |f| \leqslant \epsilon b_n + \sup_{x \in [0,1]} |f(x)| (1 - b_n)$$