Pulling limits into integrals

Question

Consider a bounded continuous function $f:[0,\infty)\to\mathbb R$ where $\lim_{t\to\infty}f(t)$ exists. Let $g:\mathbb [0,\infty)\to\mathbb R$ be another continuous function.

I'm not sure about how to justify

$$\lim_{t\to\infty}\int_0^{f(t)}g(s)ds=\int_0^{\lim_{t\to\infty}f(t)}g(s)ds$$

as a basic continuity argument breaks down as of the improper limit. Could I maybe argue as follows:

$$\lim_{t\to\infty}\int_0^{f(t)}g(s)ds=\lim_{x\to\lim_{t\to\infty}f(t)}\int_0^xg(s)ds=\int_0^{\lim_{t\to\infty}f(t)}g(s)ds$$

I'm not totally sure about this rewriting of the limit.

Answer 1

This result is definitely true intuitively, but is difficult to show by basic continuity arguments. As such, I have drafted a solution using a more traditional $\epsilon-\delta$ argument.

We will denote

$$\lim_{t\to\infty}f(t)=L$$

for convenience. Since $g$ is a continuous function, its integral is as well, so we have

$$G(t)=\int_0^tg(s)ds$$

is continuous; more importantly, it is continuous at $L$, so for any $\epsilon>0$, there exists a $\delta>0$ such that if $|t-L|<\delta$, then $|G(t)-G(L)|<\epsilon$. For this same $\delta$, there exists some $M\in\mathbb{R}$ such that for all $t>M$, $|f(t)-L|<\delta$. This means that for the given $\epsilon>0$, there exists some $M\in\mathbb{R}$ such that we have the following implications

$$t>M\implies |f(t)-L|<\delta\implies |G(f(t))-G(L)|<\epsilon$$

By definition, $\lim\limits_{t\to\infty}G(f(t))=G(L)$, or in other words,

$$\lim_{t\to\infty}\int_0^{f(t)}g(s)ds=\int_0^Lg(s)ds=\int_0^{\lim_{t\to\infty}f(t)}g(s)ds$$

Answer 2

The key is that integral is a continuous function of its upper limit. More formally we have

Fundamental Theorem of Calculus Part 1: Let the function $f:[a, b] \to\mathbb {R} $ be Riemann integrable on $[a, b] $ and let $F:[a, b] \to\mathbb{R} $ be defined by $$F(x) =\int_{a} ^{x} f(t) \, dt$$ Then $F$ is continuous on $[a, b] $ and if $f$ is continuous at some point $c\in[a, b] $ then $F$ is differentiable at $c$ with $F'(c) =f(c) $.

For the current problem consider $$G(x) =\int_{0}^{x}g(s)\,ds$$ then by the above theorem $G$ is continuous on $[0,\infty) $ and we have to deal with the limit $$\lim_{t\to\infty} \int_{0}^{f(t)}g(s)\,ds=\lim_{t\to \infty} G(f(t)) $$ Since $G$ is continuous we can switch the limits to get $$\lim_{t\to \infty} G(f(t)) =G(\lim_{t\to\infty} f(t)) $$ Note that in the above we don't use continuity of $f$ or $g$. The result is true if $g$ is Riemann integrable on any finite subinterval of $[0,\infty)$ and if $\lim_{t\to\infty} f(t) $ exists.