Instead of Dominated Convergence Theorem

Question

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?

Answer 1

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 $(B,\infty )$.

Answer 2

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.