I've got a problem.
Let $g(t)\ge0$ and it has improper integral on interval $[A, B)$. Furthermore, sequence of integrable functions $f_{n}(t)$ is uniformly convergent do $f(t)$ on every subinterval $[a,b]$ of $[A,B)$ and $0 \le f_{n}(t) \le g(t)$ for every $t\in[A,B)$
Prove that
$$\lim_{n\to\infty} \int_{A}^{B^{-}}f_{n}(t)dt = \int_{A}^{B^{-}}f(t)dt$$
As I suppose our goal is to show that the difference
$$|\int_{A}^{B^{-}}f_{n}(t)dt-\int_{A}^{B^{-}}f(t)dt| < \epsilon$$
I've tried to do it like this
$$|\int_{A}^{B^{-}}f_{n}(t)dt-\int_{A}^{B^{-}}f(t)dt| = |\int_{A}^{b}f_{n}(t)dt-\int_{A}^{b}f(t)dt + \int_{b}^{B^{-}}f_{n}(t)dt - \int_{b}^{B^{-}}f(t)dt| \le$$
$$\le \int_{A}^{b}|f_{n}(t)-f(t)|dt + |\int_{b}^{B^{-}}(f_{n}(t)-f(t))|$$
From the uniformly convergence we can make this first integral infitesimal $(b-A)\epsilon$, but don't know how to estimate the second one. As I think it has to be connected with function $g(t)$, but is it correct to say it is lower than $2\int_{b}^{B^{-}}g(t)dt$, which we can make infinitesimal, because of convergence of improper integral $\int_{A}{B^{-}}g(t)dt$.
Edit:
Okey, that's my last try
From uniform convergence of $f_{n}(t)$ to $f(t)$ We can make the difference $|f_{n}(t)-f(t)| < \epsilon_1 = \frac{\epsilon}{3(b-A)}$ then
$$\int_{A}^{b}|f_{n}(t)-f(t)|dt < (b-A)\cdot\epsilon_1 = \frac{\epsilon}{3}$$
Next step, it is known that our improper integral of $g(t)$ exists, so
$$\lim_{b\to B}\int_{b}^{B}g(t)dt = 0$$
Hence, we can make $|\int_{b}^{B}g(t)dt| < \epsilon_2 = \frac{\epsilon}{3}$
To sum up
$$\int_{A}^{b}|f_{n}(t)-f(t)|dt + |\int_{b}^{B^{-}}(f_{n}(t)-f(t))| \le (b-A)\cdot\epsilon_1 + 2\epsilon_2 = 3\cdot\frac{\epsilon}{3} = \epsilon$$