Limit of improper integrals of uniformly convergent function 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
$$\int_{A}^{b}|f_{n}(t)-f(t)|dt + |\int_{b}^{B^{-}}(f_{n}(t)-f(t))| \le (b-A)\cdot\epsilon_1 + 2\epsilon_2 = 3\cdot\frac{\epsilon}{3} = \epsilon$$
 A: I'm just cleaning up the original poster's answer, and formalizing Daniel Fischer's good hints in the comments, for anyone who wants all the details.
The limit we're trying to prove is equivalent to showing,
$\left| \int_A^{B-} f-f_n \right| < \epsilon$ for all $n$ sufficiently large.
We have
$\left| \int_A^{B-} f-f_n \right| \le \left| \int_A^{b} f-f_n \right| + \left| \int_b^{B-} f-f_n \right|$.
By the bounds on $f$ and $f_n$ with respect to $g$, we have $|f-f_n| \le g(x)$, so,
$\left| \int_b^{B-} f-f_n \right| \le \int_b^{B-} |f-f_n| \le \int_b^{B-} g < \lim_{c \uparrow B} \int_b^c g $. (The last inequality follows from $g\ge0$.)
Now choose $b$ sufficiently large such that $\int_b^{B} g < \epsilon/2$ (as guaranteed by the integrability of $g$).
With $b$ now fixed, we have for all $n$ sufficiently large that, by uniform convergence of $f_n \to f$,
$\left| \int_A^b f-f_n \right| < \epsilon/2$. We can conclude that for all $n$ sufficiently large,
$\left| \int_A^{B-} f-f_n \right| < \epsilon/2 + \epsilon/2$. Taking the limit in $n$ gives,
$\left| \int_A^{B-} f-\lim_{n \to \infty} \int_A^{B-} f_n \right| < \epsilon$. Since $\epsilon$ is arbitrary, 
$\int_A^{B-} f = \lim_{n \to \infty} \int_A^{B-} f_n$.
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(Edited slightly to include $\int_b^{B-} g < \lim_{c \uparrow B} \int_b^c g$; why the downvote? also including a comment is more helpful for the community, and for those making good faith efforts.)
