Let $A(t)$ be a continuous real-valued square matrix of size $n\times n$. Show that every solution of the nonautonomous linear system (where $Y(t) \in \mathbb R^n$ for all $t$ in the domain of $Y$) $$Y'(t) = A(t)Y(t)$$ satisfies $$|Y(t)| \le |Y(0)|\exp\left(\int_0^t \|A(s)\|\, ds\right) \tag{1}$$ Further, show that if $\int_0^\infty \|A(s)\|\, ds < \infty$, then every solution of this system has a finite limit as $t$ approaches infinity.
My work: Suppose $$|Y(t)| \le |Y(0)|\exp\left(\int_0^t \|A(s)\|\, ds\right)$$ is true; and take $t\to\infty$. Since we don't know if $\lim_{t\to\infty} Y(t)$ exists (yet?), it makes sense to work with $\liminf_{t\to\infty}$ and $\limsup_{t\to\infty}$ instead. So, we have $$\left|\liminf_{t\to\infty} Y(t)\right| \le |Y(0)|\exp\left(\int_0^\infty \|A(s)\|\, ds\right) \quad\text{and}\quad \left|\limsup_{t\to\infty} Y(t)\right| \le |Y(0)|\exp\left(\int_0^\infty \|A(s)\|\, ds\right)$$ If we can show that $\lim_{t\to\infty} Y(t)$ exists, then we are done (by any of the two inequalities above). How do we do that? Further, we also need to prove $(1)$.
Thanks!
For $t \ge 0$ we have $Y(t)=Y(0)+ \int_0^t A(s)Y(s)ds$, hence $$ |Y(t)| \le |Y(0)| + \int_0^t \|A(s)\|\cdot|Y(s)|ds. $$ Now (1) follows from the Grönwall-inequality: https://en.wikipedia.org/wiki/Gr%C3%B6nwall%27s_inequality
If in addition $\int_0^\infty \|A(s)\| ds < \infty$ we first get from (1) that $Y$ is bounded on $[0, \infty)$: $$ |Y(t)| \le |Y(0)| \exp(\int_0^\infty \|A(s)\| ds)=:\beta $$
Let $\varepsilon > 0$ and choose $t_0$ such that $\int_{t_0}^t \|A(s)\| ds < \varepsilon$ $(t \ge t_0)$. For $t>\tau \ge t_0$ we have $$ |Y(t)-Y(\tau)| \le \int_\tau^t \|A(s)\|\cdot|Y(s)|ds \le \beta \int_{t_0}^t\|A(s)\|ds \le \beta \varepsilon. $$ Thus, according to the Cauchy-criterion $\lim_{t \to \infty} Y(t)$ exists.