Let $f_n\in C(a,b)$, and $f_n\rightrightarrows f$ on $[a+\delta,b-\delta]$, $\forall\ \delta>0$. Moreover, the improper integral $\int_a^b f_n, \int_a^b f$ exists, and $\lim_{n\to\infty} \int_a^b f_n=\int_a^b f$. Show $\lim_{n\to\infty} \int_a^b |f_n-f|=0$.
What I know it only that $\lim \int_{a+\delta}^{b-\delta} f_n=f$. How to first choose such a $\delta$ such that $\int_a^{a+\delta}|f_n-f|$ is sufficiently small? Here $n$ varies infinitely.
This is false. For a counterexample, let $a=-1, b=1, f=0$ and $$f_n(x) = \begin{cases}n^2(x+1)-n, \quad & -1 < x \leq -1+\frac1n \\ 0 , & -1+\frac1n < x < 1-\frac1n \\ n^2(x-1)+n, & 1-\frac1n \leq x < 1\end{cases}.$$
Given any $\delta > 0,$ for any $n > \frac1\delta$ we have $f_n=f$ on $[-1+\delta, 1-\delta]$ (and "is eventually identical to" is stronger than "converges uniformly to").
Additionally, we may show that the improper integrals exist (the definitions extend to continuous functions on $[-1,1]$), and $\int_{-1}^1 f_n = 0 = \int_{-1}^1 f$ for all $n$, and the limit of $0$ is $0$, so each condition in the problem is satisfied.
However, $\int_{-1}^1 |f_n-f| = 1$ doesn't converge to $0$.
Note: It's possible you meant $\lim_{n\to\infty}\int_a^b |f_n| = \int_a^b |f|$ instead of $\lim_{n\to\infty} \int_a^b f_n = \int_a^b f$, in which case it should be true.