Let $f_n:(a,b)\to \mathbb{R}$ be a sequence of continuous functions such that $$|f_n(x)|\leq M\:\:\:\: \text{for all $n\geq$ 1,$\:\:$$x\in(a,b),$}$$
for some constant $M>0$, and $$\lim_{n\to{\infty}}f_n(x)=f(x)\:\:\:\text{for all $x\in(a,b)$ },$$
For some $f:(a,b)$. Moreover, for any small $\epsilon>0,$ the above convergence is uniform on $[a+\epsilon,b-\epsilon]. $ Show that $f(.)$ is continuous and $$\lim_{n\to \infty}\int_{a}^bf_n(x)dx=\int_a^bf(x)dx.$$
If I can show $f(.)$ is continuous, then the rest would be so easy. let $\epsilon>0$. As $f_n$ converges uniformly on $[a+\epsilon,b-\epsilon,]$, so $f(x)$ is continuous on $[a+\epsilon,b-\epsilon].$ Also, I can claim $ (f_n)$ is uniformly equicontinuous on $[a+\epsilon,b-\epsilon].$ However, I don't know how to show $f(x)$ is continuous on $[a,a+\epsilon] $ and $[b-\epsilon, b]$. I appreciate any hint, help or any link.
Hint. As regards $\lim_{n\to \infty}\int_{a}^bf_n(x)dx=\int_a^bf(x)dx$, note that $$\left|\int_{a}^{b}(f_n(x)-f(x))dx \right|= \left|\int_{a}^{a+\epsilon} \right|+\left|\int_{a+\epsilon}^{b-\epsilon} \right|+ \left|\int_{b-\epsilon}^{b} \right|\leq 2M\epsilon+\epsilon +2M\epsilon$$ where for the second integral we use uniform convergence and for the first and the third integral we use the the uniform boundedness.