Spectral Measures: Core Lemma Given a Hilbert space $\mathcal{H}$.
Consider a Hamiltonian:
$$H:\mathcal{D}(H)\to\mathcal{H}:\quad H=H^*$$
Regard a dense domain:
$$\mathcal{D}\subseteq\mathcal{D}(H):\quad\overline{\mathcal{D}}=\mathcal{H}$$
The core lemma says:
$$e^{itH}\mathcal{D}\subseteq\mathcal{D}\implies\overline{H_\mathcal{D}}=H$$
How can I prove this from scratch?
 A: A symmetric operator $A$ is densely-defined and selfadjoint iff $(A \pm i I)$ are surjective. The operator $A$ has a densely-defined, selfadjoint closure iff $(A\pm iI)$ have dense ranges.
In your case, you want to show that the restriction $H_{\mathcal{D}}$ of the densely-defined selfadjoint operator $H$ to $\mathcal{D}$ has closure equal to $H$. Because $H_{\mathcal{D}}$ is symmetric on its domain, then this is equivalent to showing that $(H_{\mathcal{D}}\pm iI)$ have dense ranges.

Claim: Under the stated assumptions, $\overline{\mathcal{D}}=\overline{\mathcal{R}(H_{\mathcal{D}}\pm iI)}$

Suppose $x \perp \mathcal{R}(H_{\mathcal{D}}+iI)$. Then
$$
           ((H+iI)h,x)=0,\;\;\; h\in\mathcal{D}.
$$
By assumption, $e^{itH}h \in \mathcal{D}$ for all $t$ whenever $h\in\mathcal{D}$. Hence, also,
$$
               ((H+iI)e^{itH}h,x)=0,\\
              \frac{d}{dt}(e^{-t}e^{itH}h,x)=0 \\
        \implies (e^{itH}h,x)=e^{t}(h,x),\;\;\; t \in\mathbb{R}.
$$
However, $(e^{itH}h,x)$ is uniformly bounded for all $t$ by $\|h\|\|x\|$ and, thus,
$(h,x)=0$ for all $h\in\mathcal{D}$. The same type of argument works for $x\perp \mathcal{R}(H_{\mathcal{D}}-iI)$. So,
$$
  \mathcal{R}(H_{\mathcal{D}}\pm iI)^{\perp} \subseteq \mathcal{D}^{\perp} \\
  \overline{\mathcal{D}}\subseteq\overline{\mathcal{R}(H_{\mathcal{D}}\pm iI)}.
$$
Conversely, if $x \in \mathcal{D}^{\perp}$, then the invariance of $\mathcal{D}$ under $e^{itH}$ gives
$$
     e^{\pm t}(e^{itH}h,x) =0,\;\;\; t \in\mathbb{R},\; h\in\mathcal{D} \\
           \implies ((H_{\mathcal{D}}\pm iI)h,x) = 0,\;\;\; h\in\mathcal{D} \\
           \implies x \in \mathcal{R}(H_{\mathcal{D}}\pm iI)^{\perp}.
$$
Therefore,
$$
          \mathcal{D}^{\perp}\subseteq \mathcal{R}(H_{\mathcal{D}}\pm iI)^{\perp} \\
        \overline{\mathcal{R}(H_{\mathcal{D}}\pm iI)}\subseteq\overline{\mathcal{D}}. \;\;\;\blacksquare
$$

Conclusion: $\mathcal{D}$ is a core for $H$ iff $\mathcal{D}$ is a dense subspace of $\mathcal{H}$.

Because you assume $\mathcal{D}$ is dense, then $\mathcal{D}$ is a core for $H$.
