Consider $[0,1] $ with $\mu$ Lebsgue measure on $[0,1]$.
Assume $H = L^2([0,1], \mu)$. Then for $\phi \in L^\infty([0,1],\mu)$, define $M_\phi \in B(L^2([0,1], \mu))$ by $$M_\phi(f) = \phi f, \ f\in L^2([0,1], \mu).$$
How can we show that $\sigma(M_\phi) = \text{ess range} \ \phi$?
A point $z\in\mathbb{C}$ is in the essential range of $\phi$ if, for every $\epsilon > 0$, one has $$ \mu \{ x \in [0,1] : |\phi(x)-z| < \epsilon \} > 0. $$ To show that such a $z$ is in the spectrum of $M_{\phi}$, let $$ f_{z,\epsilon} = \chi_{\{ x \in [0,1] : |\phi(x)-z| < \epsilon \}}. $$ Then, $\|f_{z,\epsilon}\| \ne 0$, and $$ \|(M_\phi-zI)f_{z,\epsilon}\|^2 = \int_{\{ x\in [0,1] : |\phi(x)-z|<\epsilon \}}|\phi(x)-z|^2dx \le \epsilon^2\|f_{z,\epsilon}\|^2. $$ That's enough to show that $M_{\phi}-zI$ cannot have a bounded inverse. So $z\in\sigma(M_{\phi})$ whenever $z$ is in the essential range of $\phi$. If $z$ is not in the essential range, then $M_{\phi}-zI$ does have a bounded inverse given as multiplication by $\frac{1}{\phi(x)-z}$.