Given a selfadjoint Operator on some seperable $\mathbb{C}$-Hilbert space $T:H\supseteq D(T) \rightarrow H$, $T$ posses a spectral representation of the form $T=U^{-1} M_{f} U$ where $f: \Omega \rightarrow \mathbb{R}$ is Borel measurable on some $\sigma$-finite measure space $(\Omega,\mathcal{A},\mu)$, and $U: H \rightarrow L^2(\Omega,\mathcal{A},\mu)$ is a unitary operator. $M_f: L^2(\Omega,\mathcal{A},\mu)\supseteq UD(T) = D(M_f)\rightarrow L^2(\Omega,\mathcal{A},\mu),g\mapsto fg$ is the multiplikation Operator.
The spectral measure is defined as $P_{\Omega}:=U^{-1} M_{\chi_{f^{-1}(\Omega)}} U$, for some Borel measurable $\Omega\subseteq \mathbb{R}$. $P_\Omega$ is a conitnous linear Operator from $H$ to $H$.
Now my question would be: How can I prove that $P_{\Omega}H\subseteq H$ is closed?
Actually I think its wrong what I stated in my last comment. $P_\Omega^2=P_\Omega$ would be enough if $P_\Omega$ would be injective which must not be the case - I apologize for that. But if $P_\Omega=P_\Omega^*$ then closedness follows like you mentioned.
This follwos by:
Let $x\in \mathrm{ker}(I-P_\Omega)$ then $(I-P_\Omega )x=0$, hence $x\in \mathrm{ran}P_\Omega$. Therefore
$\mathrm{ran}{P_\Omega}\supseteq \mathrm{ker}(P_\Omega-I)$.
If $y\in \mathscr{H}$ is arbitrary and $x:=P_\Omega y$. Then $P_\Omega x=P_\Omega^2 y=P_\Omega y$ and thus $P_\Omega (x-y)=0$. Furthermore we have $\mathrm{ker}P_\Omega=\mathrm{ran}P_\Omega^\perp$, since $P_\Omega$ is selfadjoint. Therefore $\langle z,x-y\rangle=0$ for all $z\in \mathrm{ran}P_\Omega$. Since $x=P_\Omega y$ we get $\langle z,(P_\Omega-I)y\rangle=0$ for all $z\in \mathrm{ran}P_\Omega$. Therefore we have $(I-P_\Omega)y\in \mathrm{ran}(P_\Omega)^\perp$. This implies $\mathrm{ran}(I-P_\Omega)\subseteq \mathrm{ran}(P_\Omega)^\perp=\mathrm{ker}(P_\Omega)$ which is equivalent to $\mathrm{ker}(P_\Omega)^\perp=\overline{\mathrm{ran}{P_\Omega}}\subseteq \mathrm{ker}(P_\Omega-I)$.
Hence
$\mathrm{ran}{P_\Omega}\subseteq \mathrm{ker}(P_\Omega-I)$.
The selfadjointnes of $P_\Omega$ follows by $(U^{-1}M_{g} U)^*=U^{-1}M_{\overline{g}} U$.