[proofplan] The proof has three independent parts. For density, we average the orbit by setting $x_h := \frac{1}{h}\int_0^h T(s)x \, d\mathcal{L}^1(s)$; the semigroup law lets us compute $\frac{1}{r}(T(r) - I)x_h$ explicitly via a telescoping integral, showing $x_h \in D(A)$, while strong continuity at $0$ gives $x_h \to x$. For closedness, we differentiate the identity $T(r)x_n - x_n = \int_0^r T(s) A x_n \, d\mathcal{L}^1(s)$, pass to the limit using uniform boundedness of $\{T(s)\}$ on $[0, r]$, and divide by $r$. For the differentiation formula, we exploit the commutativity $T(t)T(h) = T(h)T(t)$ to compute the right and left derivatives of $T(t)x$ separately. Throughout we use the exponential bound $\|T(t)\|_{\mathcal{L}(X)} \le M e^{\omega t}$ which holds for every $C_0$-semigroup. [/proofplan]

[step:Fix the standing exponential bound and Bochner-integral framework] By the boundedness theorem for $C_0$-semigroups, there exist constants $M \ge 1$ and $\omega \in \mathbb{R}$ such that \begin{align*} \|T(s)\|_{\mathcal{L}(X)} \le M e^{\omega s} \quad \text{for all } s \ge 0. \end{align*} Combined with the strong continuity $s \mapsto T(s)x$ from $[0, \infty)$ to $X$, the map $s \mapsto T(s)x$ is continuous and locally bounded for each $x \in X$, so the Bochner integral \begin{align*} \int_a^b T(s)x \, d\mathcal{L}^1(s) \in X \end{align*} is well-defined for every $0 \le a \le b < \infty$ and every $x \in X$. We use this integral repeatedly below. Recall that the generator is defined by \begin{align*} D(A) &:= \left\{ x \in X : \lim_{h \to 0^+} \frac{T(h)x - x}{h} \text{ exists in } X \right\}, & Ax &:= \lim_{h \to 0^+} \frac{T(h)x - x}{h}. \end{align*} [/step]

[step:Show that the orbit average $x_h := \frac{1}{h}\int_0^h T(s)x \, d\mathcal{L}^1(s)$ lies in $D(A)$][claim:For every $x \in X$ and $h > 0$, the element $x_h := \frac{1}{h}\int_0^h T(s)x\, d\mathcal{L}^1(s)$ belongs to $D(A)$, and $A x_h = \frac{1}{h}(T(h) - I)x$.] [/claim] [proof] Fix $x \in X$, $h > 0$, and let $r \in (0, h)$. Using the semigroup property $T(r)T(s) = T(r + s)$ together with the fact that the bounded operator $T(r)$ commutes with the Bochner integral, we compute \begin{align*} T(r) x_h &= \frac{1}{h} T(r) \int_0^h T(s) x \, d\mathcal{L}^1(s) = \frac{1}{h} \int_0^h T(r + s) x \, d\mathcal{L}^1(s). \end{align*} Substituting $u = r + s$ (with $d\mathcal{L}^1(s) = d\mathcal{L}^1(u)$ and the domain $s \in [0, h]$ mapping to $u \in [r, r + h]$): \begin{align*} T(r) x_h &= \frac{1}{h} \int_r^{r+h} T(u) x \, d\mathcal{L}^1(u). \end{align*} Subtracting $x_h = \frac{1}{h}\int_0^h T(u) x\, d\mathcal{L}^1(u)$ and cancelling the overlapping interval $[r, h]$ (valid for $r < h$): \begin{align*} T(r) x_h - x_h &= \frac{1}{h}\left( \int_h^{r+h} T(u) x \, d\mathcal{L}^1(u) - \int_0^r T(u) x \, d\mathcal{L}^1(u) \right). \end{align*} Dividing by $r$: \begin{align*} \frac{T(r) x_h - x_h}{r} &= \frac{1}{h}\left( \frac{1}{r}\int_h^{r+h} T(u) x\, d\mathcal{L}^1(u) - \frac{1}{r}\int_0^r T(u) x\, d\mathcal{L}^1(u) \right). \end{align*} Since $u \mapsto T(u)x$ is continuous on $0, \infty)$, the [Lebesgue Differentiation Theorem for vector-valued continuous integrands gives \begin{align*} \lim_{r \to 0^+} \frac{1}{r}\int_h^{r+h} T(u)x\, d\mathcal{L}^1(u) &= T(h)x, & \lim_{r \to 0^+} \frac{1}{r}\int_0^r T(u)x\, d\mathcal{L}^1(u) &= T(0)x = x. \end{align*} Therefore the limit defining $A x_h$ exists, $x_h \in D(A)$, and \begin{align*} A x_h &= \frac{1}{h}(T(h)x - x). \end{align*} [/proof][/step]

[guided]Fix $x \in X$ and $h > 0$. We want to show that the average $x_h := \frac{1}{h}\int_0^h T(s)x\, d\mathcal{L}^1(s)$ is in the domain of $A$, and that we can compute $Ax_h$ explicitly. To do this, we must evaluate the limit \begin{align*} \lim_{r \to 0^+} \frac{T(r)x_h - x_h}{r} \end{align*} and show it exists in $X$. The strategy is to apply $T(r)$ inside the integral and use the semigroup property to convert $T(r)T(s) = T(r + s)$ into a translation of the integration variable; the resulting expression then becomes a difference of two integrals over short intervals, whose limit we recognise via the fundamental theorem of calculus for Bochner integrals. Let $r \in (0, h)$. The bounded operator $T(r) \in \mathcal{L}(X)$ commutes with the Bochner integral, and the semigroup law gives \begin{align*} T(r) x_h &= \frac{1}{h} T(r) \int_0^h T(s) x \, d\mathcal{L}^1(s) = \frac{1}{h} \int_0^h T(r + s) x \, d\mathcal{L}^1(s). \end{align*} We change variables: let $u = r + s$, so $d\mathcal{L}^1(s) = d\mathcal{L}^1(u)$ and as $s$ ranges over $[0, h]$ the new variable $u$ ranges over $[r, r + h]$. This gives \begin{align*} T(r) x_h &= \frac{1}{h} \int_r^{r+h} T(u) x \, d\mathcal{L}^1(u). \end{align*} Now we subtract $x_h = \frac{1}{h}\int_0^h T(u) x\, d\mathcal{L}^1(u)$. The two integrals overlap on $[r, h]$ (here is where we use $r < h$, so the interval is non-empty), and the overlap cancels: \begin{align*} T(r) x_h - x_h &= \frac{1}{h}\left( \int_h^{r+h} T(u) x \, d\mathcal{L}^1(u) - \int_0^r T(u) x \, d\mathcal{L}^1(u) \right). \end{align*} Dividing by $r$ rearranges this into a sum of two short-interval averages: \begin{align*} \frac{T(r) x_h - x_h}{r} &= \frac{1}{h}\left( \frac{1}{r}\int_h^{r+h} T(u) x\, d\mathcal{L}^1(u) - \frac{1}{r}\int_0^r T(u) x\, d\mathcal{L}^1(u) \right). \end{align*} Each of these is the average of a continuous function $u \mapsto T(u)x$ over a small interval. Since $u \mapsto T(u)x$ is continuous on $[0, \infty)$ (the strong continuity hypothesis), the fundamental theorem of calculus for Bochner integrals yields \begin{align*} \lim_{r \to 0^+} \frac{1}{r}\int_h^{r+h} T(u)x\, d\mathcal{L}^1(u) &= T(h)x, & \lim_{r \to 0^+} \frac{1}{r}\int_0^r T(u)x\, d\mathcal{L}^1(u) &= T(0)x = x. \end{align*} Therefore the limit defining $A x_h$ exists, which by definition means $x_h \in D(A)$, and \begin{align*} A x_h &= \frac{1}{h}(T(h)x - x). \end{align*} The key idea is that averaging the orbit smooths it: even when $x \notin D(A)$, the average $x_h$ lies in $D(A)$ because the integral can absorb a derivative via the fundamental theorem of calculus.[/guided]

[step:Pass $h \to 0^+$ to deduce density of $D(A)$] Strong continuity at $0$ states that $\lim_{s \to 0^+} T(s)x = T(0)x = x$ in $X$. By the Lebesgue Differentiation Theorem for vector-valued continuous integrands, \begin{align*} \lim_{h \to 0^+} x_h &= \lim_{h \to 0^+} \frac{1}{h} \int_0^h T(s) x \, d\mathcal{L}^1(s) = T(0) x = x. \end{align*} By the previous step, $x_h \in D(A)$ for every $h > 0$. Hence $x$ is a limit of elements of $D(A)$, so $x \in \overline{D(A)}$. Since $x \in X$ was arbitrary, $\overline{D(A)} = X$, i.e. $D(A)$ is dense in $X$. [/step]

[step:Establish the integral identity $T(r)x - x = \int_0^r T(s) Ax \, d\mathcal{L}^1(s)$ for $x \in D(A)$] [claim:For $x \in D(A)$ and $r \ge 0$, \begin{align*} T(r)x - x &= \int_0^r T(s) Ax \, d\mathcal{L}^1(s). \end{align*}] [/claim] [proof] Fix $x \in D(A)$ and consider $\varphi: [0, \infty) \to X$, $s \mapsto T(s)x$. We show $\varphi$ is differentiable on $[0, \infty)$ with $\varphi'(s) = T(s)Ax$, then integrate. For $h > 0$ and $s \ge 0$, the semigroup law $T(s + h) = T(s) T(h) = T(h)T(s)$ gives \begin{align*} \frac{T(s + h)x - T(s)x}{h} &= T(s)\, \frac{T(h)x - x}{h}. \end{align*} Since $x \in D(A)$, $\frac{T(h)x - x}{h} \to Ax$ as $h \to 0^+$, and $T(s) \in \mathcal{L}(X)$ is continuous, so the right-hand side converges to $T(s)Ax$. Hence $\varphi$ has right derivative $T(s)Ax$ at every $s \ge 0$. For $s > 0$ and $0 < h < s$, write \begin{align*} \frac{T(s)x - T(s - h)x}{h} - T(s) Ax &= T(s - h)\left( \frac{T(h)x - x}{h} - Ax \right) + (T(s - h) - T(s)) Ax. \end{align*} The first term satisfies, using $\|T(s - h)\|_{\mathcal{L}(X)} \le M e^{\omega s}$ (uniform on the compact interval $[0, s]$), \begin{align*} \left\| T(s - h)\left( \frac{T(h)x - x}{h} - Ax \right) \right\|_X &\le M e^{|\omega| s}\, \left\| \frac{T(h)x - x}{h} - Ax \right\|_X \xrightarrow{h \to 0^+} 0. \end{align*} The second term tends to $0$ by strong continuity: $T(s - h) Ax \to T(s) Ax$. Hence $\varphi$ has left derivative $T(s)Ax$ at every $s > 0$. The right derivative on $[0, \infty)$ and the left derivative on $(0, \infty)$ agree, so $\varphi \in C^1([0, \infty); X)$ with $\varphi'(s) = T(s) Ax$. Since $s \mapsto T(s) Ax$ is continuous and hence Bochner-integrable on $[0, r]$, the Fundamental Theorem of Calculus for Bochner Integrals gives \begin{align*} T(r)x - x &= \varphi(r) - \varphi(0) = \int_0^r T(s) Ax \, d\mathcal{L}^1(s). \end{align*} [/proof] [/step]

[step:Deduce closedness of $A$ from the integral identity] Suppose $x_n \in D(A)$ with $x_n \to x$ and $Ax_n \to y$ in $X$. We show $x \in D(A)$ and $Ax = y$. By the previous step, for every $r > 0$ and every $n$, \begin{align*} T(r) x_n - x_n &= \int_0^r T(s) A x_n \, d\mathcal{L}^1(s). \end{align*} On the left, $T(r) x_n - x_n \to T(r) x - x$ since $T(r) \in \mathcal{L}(X)$ is continuous. On the right, the integrand $T(s) A x_n$ converges to $T(s) y$ for each $s \in [0, r]$, with the uniform bound \begin{align*} \|T(s)(Ax_n - y)\|_X &\le M e^{|\omega| r} \|Ax_n - y\|_X \xrightarrow{n \to \infty} 0, \end{align*} i.e. uniform convergence on $[0, r]$. Therefore the Bochner integral converges: \begin{align*} \int_0^r T(s) A x_n \, d\mathcal{L}^1(s) &\xrightarrow{n \to \infty} \int_0^r T(s) y \, d\mathcal{L}^1(s). \end{align*} Passing to the limit in the integral identity yields, for every $r > 0$, \begin{align*} T(r) x - x &= \int_0^r T(s) y \, d\mathcal{L}^1(s). \end{align*} Dividing by $r$ and letting $r \to 0^+$, the right-hand side tends to $T(0)y = y$ by the Lebesgue Differentiation Theorem (using continuity of $s \mapsto T(s)y$). Hence \begin{align*} \lim_{r \to 0^+} \frac{T(r) x - x}{r} &= y, \end{align*} which is exactly the statement that $x \in D(A)$ and $Ax = y$. This proves $A$ is closed. [/step]

[step:Establish the differentiation formula $\frac{d}{dt}T(t)x = AT(t)x = T(t)Ax$ for $x \in D(A)$] Fix $x \in D(A)$ and $t \ge 0$. Step 4 already showed that $\varphi: s \mapsto T(s)x$ is in $C^1([0, \infty); X)$ with right derivative $T(s)Ax$ at every $s \ge 0$ and left derivative $T(s)Ax$ at every $s > 0$. So \begin{align*} \frac{d}{dt} T(t) x &= T(t) Ax. \end{align*} It remains to show that $T(t) x \in D(A)$ and $A T(t) x = T(t) A x$. For $h > 0$, by the semigroup property and continuity of $T(t)$: \begin{align*} \frac{T(h)\, T(t) x - T(t) x}{h} &= \frac{T(t + h) x - T(t) x}{h} = T(t)\, \frac{T(h) x - x}{h}. \end{align*} The right-hand side converges to $T(t) A x$ as $h \to 0^+$, since $T(t) \in \mathcal{L}(X)$ is continuous and $\frac{T(h) x - x}{h} \to A x$. Hence the left-hand side converges, which by definition of $D(A)$ means $T(t) x \in D(A)$ and \begin{align*} A T(t) x &= T(t) A x. \end{align*} Combining with $\frac{d}{dt} T(t) x = T(t) Ax$: \begin{align*} \frac{d}{dt} T(t) x &= A T(t) x = T(t) A x. \end{align*} This completes the proof of all three assertions. [/step]