Instead of Dominated Convergence Theorem It is a traditional result that 
\[
\lim_{n \to \infty} \int_{A}^{\infty} f_{n}(t) dt
= \int_{A}^{\infty} f(t) dt
\]
if $f_{n}(t) \to f(t)$ for each $t$ and there exists an integrable function $\phi(t)$ 
satisfying
\[
|f_{n}(t)| \leq \phi(t)
\]
for all sufficiently large $n$.
On the other hands, as a undergrad student, I remember reading
in a book by T.M. Apostol that
the uniform convergence of integrals can somehow replace
the domination condition; uniform convergence of improper
integrals is defined as follows; given $\epsilon > 0$,
there exists a $B$ such that whenever $B_{1}, B_{2} > B$,
\[
|\int_{B_{1}}^{B_{2}}f_{n}(t)dt | < \epsilon
\]
for all $n$.
Is my memory correct?
 A: Certainly! If $f_n \to f$ uniformly and the improper integrals converge uniformly then $\int_A ^{\infty } f_n (t) dt \to \int_A ^{\infty } f (t) dt$ and the proof is very simple. You just have to split the integrals into those over $(A,B)$ and $(B,\infty )$.
A: Without uniform convergence of the improper integral, we have the counterexample $f_n(x) = x^{-1} \chi_{[1,n]}(x)$ where
$$\int_1^\infty f_n = \int_1^n x^{-1} \, dx = \log n$$
is convergent and $f_n(x) \to x^{-1}$ uniformly on $[1,\infty)$, but $x^{-1}$ is not integrable on $[1,\infty)$.
With the stronger condition of uniform convergence (both for the improper integrals and $f_n \to f$) we can be assured that $f$ is integrable on any closed and bounded interval $[a,b]$ and that
$$\lim_{n \to \infty}\int_{a}^{b} f_n  = \int_{a}^{b} f$$ 
We then can show that $\int_0^\infty f$ converges.  For any $\epsilon >0$ there exists $B$ (which may depend on $\epsilon$, but not on $n$) such that if $b_2 > b_1 > B$, then we have
$$\left|\int_{b_1}^{b_2} f \right| = \lim_{n \to \infty}\left|\int_{b_1}^{b_2} f_n \right| \leqslant \epsilon$$
implying that the improper integral $\int_a^\infty f$ converges by the Cauchy criterion.
There exists $c$ (independent of $n$) such that
$$\left|\int_c^\infty f_n\right| , \, \left|\int_c ^\infty f\right| < \frac{\epsilon}{3}$$
We also have 
$$\left|\int_a^\infty f_n - \int_a^\infty f\right| \leqslant \left|\int_a^c f_n - \int_a^c f\right| + \left|\int_c^\infty f_n\right| + \left|\int_c ^\infty f\right| \\ \leqslant   \left|\int_a^c f_n - \int_a^c f\right|  + \frac{2\epsilon}{3}$$
Given the uniform convergence $f_n \to f$ on the bounded interval $[a,c]$ we can show that the first term on the RHS is less than $\epsilon/3$ for sufficiently large $n$ and conclude.
