Every solution $x(t)$ of $x′(t)=A(t)x(t)$ converges, question about the answer

Question

So in the post: Show that every solution $x(t)$ of $x'(t)= A(t)x(t)$ converges to some limit

(Long-time asymptotics). Suppose

$$\int_0^∞\|A(t)\|\,dt < ∞.$$

Show that every solution $x(t)$ of $x'(t)= A(t)x(t)$ converges to some limit: $\lim_{t→∞} x(t) = x_∞.$

And one of the answers says:

Once you know that $x$ is bounded by some constant $K$ you can use the integral equation to show that $x(t)$ is Cauchy, for $t_1 \leq t_2$ you have: $$\|x(t_1)-x(t_2)\| =\left\|\int_{t_1}^{t_2} A(s) x(s) ds \right\| \leq K \int_{t_1} ^{t_2} \|A(s)\| ds$$ And $\int_{t_1} ^{t_2} \|A(s)\|ds$ goes to $0$ as $t_1$ goes to $+\infty$.

QUESTION: Why does this integral goes to $0$ as $t_1$ goes to $+ \infty.$

Answer 1

By definition $\int_0^∞\|A(t)\|\,dt=\lim_{s\to\infty}\int_{0}^{s}\|A(t)\|\,dt $. Let $a(s):=\int_{0}^{s}\|A(t)\|\,dt$. Then the existence of $\lim_{s\to\infty}a(s)$ implies that for all $\varepsilon>0$ there is some $s_\varepsilon$ such that $\vert a(t_2)-a(t_1)\vert <\varepsilon$ for all $t_1,t_2>s_\varepsilon$. Finally note that $a(t_2)-a(t_1)=\int_{t_1}^{t_2} \|A(t)\|$.