Let $\mu$ be a $\sigma$-finite measure on $(\Bbb{R},\mathcal{B}(\Bbb{R}))$. Let us define an operator $T^{\mu}$ on $H=L^2(\mu)$ as follows: $$D(T^{\mu})=\{u\in H: \int (1+x^2)|u(x)|^2\ d\mu(x )<\infty\}\text{ and }(T^{\mu}u)(x)=xu(x)$$
I to show that $T^{\mu}$ is closed and densely defined. I'm able to prove that $T^{\mu}$ is closed. But I'm stuck at the density of $D(T^{\mu})$. My initial guess is to show that simple function are inside $D(T^{\mu})$. But if $E\in\mathcal{B}(\Bbb{R})$ with $\mu(E)<\infty$, $\int\limits_Ex^2\ d\mu(x)$ may not be finite.
If $\mu$ is regular measure, then $C_c$ functions are dense in $L^2$ and it is true that $C_c$ functions are in $D(T^{\mu})$. But what about general $\sigma$-finite measures?
Can anyone help me with an idea to finish the proof? Thanks for your help in advance.
Even $\sigma-$ finiteness is not necessary.
If $u \in L^{2}(\mu)$ then $u_n=u1_{|x|\leq n} \in D(T^{\mu})$ and $u_n \to u$ in $L^{2}(\mu)$.