Consider the multiplication operator by the idependent variable on the space $L_2(R)$. It is defined by
$(Mu)(t) = tu(t), t \in R$ a.e.
$dom(M)$ consists of all measurable functions $u$ which satisfy
$$ \int_{R}(1+t^2)|u(t)|^2dt < \infty $$ Show that $\sigma _c =\mathbb{R}$.
Let $t \neq t_0$ and $$g_{\varepsilon} = \left\{ \begin{array}{ll} 1 & \textrm{gdy $|t-t_0|< \varepsilon $}\\ 0 & \textrm{gdy $|t-t_0| \ge \varepsilon$} \end{array} \right.$$ $$\int_{\mathbb{R}}(t-\lambda)fg_{\varepsilon}dx=\int_{|t-t_0|<\varepsilon}(t-\lambda)fdx \xrightarrow{\varepsilon \to 0}0$$ It follows that $g_\varepsilon=0$, because $ran(t-\lambda)$ is dense but I don't know how it's imply that $\sigma_c=\mathbb{R}$
To show that the continuous spectrum is $\Bbb R$, you want to show that $M-\lambda I$ has dense range. Meaning that if $g$ is such that $\langle (M-\lambda I) f, g\rangle=0$ for all $f$, then $g\equiv 0$.
However, $\langle (M-\lambda I) f, g\rangle = \langle (t-\lambda) f, g\rangle$. Pick a delta sequence that is centered around $x_0$ for $x_0\neq \lambda$.
I think this should help.