Find projection-valued measure associated with parity operator Let's define parity operator as follows: $$\pi:L^2(\mathbb{R})\to L^2(\mathbb{R})$$ $$\psi(x)\mapsto \psi(-x)$$
It's easy to show that $\pi$ is a self-adjoint operator and its spectrum is just $\sigma(\pi)=\{-1,+1\}$. According to spectral theorem there is a unique projection-valued measure $P_{\pi}$ such that: $$\pi=\int_{\mathbb{R}}\lambda  \,dP_{\pi}(\lambda)$$
How do I find explicitly $P_{\pi}$?  

Observe for any $\psi\in L^2(\mathbb{R})$ we see that
\begin{align}
\psi(x) = \frac{\psi(x)+\psi(-x)}{2}+ \frac{\psi(x)-\psi(-x)}{2}=: \psi_\text{even}(x)+\psi_\text{odd}(x)
\end{align}
then
\begin{align}
\pi(\psi)(x) =  \psi_\text{even}(-x)+\psi_\text{odd}(-x)=\psi_\text{even}(x)-\psi_\text{odd}(x).
\end{align}
In short, we have that
\begin{align}
\pi = P_\text{even}-P_\text{odd}
\end{align}
where $P_\text{even}\psi = \psi_\text{even}$ and $P_\text{odd}\psi=\psi_\text{odd}$.
Edit: Note that
\begin{align}
P_\pi(\lambda) = \delta(\lambda-1)P_\text{even}+\delta(\lambda+1)P_\text{odd}
\end{align}
i.e. $P_\pi:\mathcal{B}_\mathbb{R}\rightarrow \mathcal{L}(L^2(\mathbb{R}),L^2(\mathbb{R}))$. Hence
\begin{align}
\int_{\mathbb{R}}\lambda\cdot dP_\pi(\lambda) = P_\text{even}-P_\text{odd}.
\end{align}
