Strongly Continuous Semigroup consequence I started studying theory of Semigroups recently and I was trying to figure out basic concepts. I know that
A semigroup $S(t)$ of bounded linear operators on $X$ is called strongly continuous ($\mathcal(C_0)$ if 
\begin{equation}
\displaystyle lim_{t \rightarrow 0} S(t)u = u
\end{equation}
 for all $u \in X$
and as a consequence I have that
\begin{equation}
lim_{h \rightarrow 0} \frac{1}{h} \int_t^{t+h}S(s)uds = S(t)u \hspace{5mm} \forall u \in X
\end{equation}
But why does this hold? thank you
 A: For all $t$, and all $u$
$$\lim_{h\to 0}S(t+h)u=\lim_{h\to 0}S(t)S(h)u=S(t)\lim_{h\to 0}S(h)u=S(t)u,$$
because $S(t)$ is continuous. If $h$ is small, then for all $s$ in $[t,t+h]$ we have
$$S(s)u\approx S(t)u$$
(the approximation in norm), so the averages of both sides over $[t,t+h]$ are also close:
$$\frac{1}{h}\int_t^{t+h}S(s)uds\approx S(t)u.$$
To formalize this, use subadditivity of integration and the fact that $S(t)u=\frac{1}{h}\int_t^{t+h}S(t)uds$.

More details for the last part: Let $\epsilon>0$. Choose $\delta>0$ such that $|h|<\delta$ implies $\Vert S(h)u-u\Vert<\epsilon/(\Vert S(t)\Vert+1)$. Then if $s\in[t,t+\delta]$, we have
$$\Vert S(s)u-S(t)u\Vert=\Vert S(t)(S(s-t)u-u)\Vert\leq\Vert S(t)\Vert \Vert S(s-t)u-u\Vert\leq\epsilon,$$
because $|s-t|<\delta$.
Let $|h|<\delta$. If we integrate the constant function $s\mapsto S(t)u$, we have
$$\int_t^{t+h}S(t)uds=h S(t)u$$
so
\begin{align*}
\left|\int_t^{t+h}S(s)uds-hS(t)u \right|
    &=\left|\int_t^{t+h}(S(s)u-S(t)u)ds\right|\\
    &\leq\int_t^{t+h}\Vert S(s)u-S(t)u\Vert ds\\
    &\leq\int_t^{t+h}\epsilon ds\\
    &=\epsilon h,
\end{align*}
where the first inequality is the subadditivity of the integral. This means that for all $|h|<\delta$,
$$\left|\frac{1}{h}\int_t^{t+h}S(s)uds-S(t)u\right|<\epsilon.$$
This is precisely what it means to have $\lim_{h\to 0}\frac{1}{h}\int_t^{t+h}S(s)uds=S(t)u$.
