[proofplan]
We prove the spectral equality by separating the two possible positions of $\lambda\in\mathbb C$. If $\lambda$ is outside the essential range, then $|\varphi-\lambda|$ is bounded below away from zero outside a null set, so multiplication by the reciprocal gives a bounded inverse for $M_\varphi-\lambda I_H$. If $\lambda$ is inside the essential range, semifiniteness supplies finite positive measure subsets on which $|\varphi-\lambda|$ is arbitrarily small, and normalized indicators of these subsets form approximate eigenvectors. The point spectrum statement follows by solving the equation $(\varphi-\lambda)f=0$ in $L^2$ and using semifiniteness once more to build a nonzero indicator eigenfunction from a positive-measure level set.
[/proofplan]
[step:Invert $M_\varphi-\lambda I$ outside the essential range]
Fix $\lambda\in\mathbb C\setminus\operatorname{ess\,ran}(\varphi)$. By the definition of essential range, there exists $\varepsilon>0$ such that, for the measurable set
\begin{align*}
A_\varepsilon:=\{\omega\in\Omega:|\varphi(\omega)-\lambda|<\varepsilon\},
\end{align*}
one has $\mu(A_\varepsilon)=0$.
Define the measurable function $g:\Omega\to\mathbb C$ as follows: for $\omega\in\Omega\setminus A_\varepsilon$, set $g(\omega)=(\varphi(\omega)-\lambda)^{-1}$, and for $\omega\in A_\varepsilon$, set $g(\omega)=0$.
For $\omega\in\Omega\setminus A_\varepsilon$ we have $|g(\omega)|\le \varepsilon^{-1}$, while $g=0$ on $A_\varepsilon$. Hence $g\in L^\infty(\Omega,\mathcal F,\mu)$ and
\begin{align*}
\|g\|_{L^\infty(\Omega)}\le \varepsilon^{-1}.
\end{align*}
Let $M_g\in\mathcal L(H)$ be the bounded multiplication operator $M_g f=gf$ for $f\in H$.
For every $f\in H$, the equality
\begin{align*}
g(\varphi-\lambda)f=f
\end{align*}
holds $\mu$-a.e., because $g(\varphi-\lambda)=1$ on $\Omega\setminus A_\varepsilon$ and $A_\varepsilon$ is null. Similarly,
\begin{align*}
(\varphi-\lambda)gf=f
\end{align*}
holds $\mu$-a.e. Therefore
\begin{align*}
M_g(M_\varphi-\lambda I_H)=I_H
\end{align*}
and
\begin{align*}
(M_\varphi-\lambda I_H)M_g=I_H
\end{align*}
as operators on $H$. Thus $M_\varphi-\lambda I_H$ is boundedly two-sided invertible in $\mathcal L(H)$, so by the definition of the resolvent set $\lambda\in\rho(M_\varphi)$. Consequently,
\begin{align*}
\sigma(M_\varphi)\subseteq \operatorname{ess\,ran}(\varphi).
\end{align*}
[guided]
Fix $\lambda\in\mathbb C\setminus\operatorname{ess\,ran}(\varphi)$. Being outside the essential range means that the values of $\varphi$ do not approach $\lambda$ on sets of positive measure. More precisely, there exists $\varepsilon>0$ such that the measurable set
\begin{align*}
A_\varepsilon:=\{\omega\in\Omega:|\varphi(\omega)-\lambda|<\varepsilon\}
\end{align*}
satisfies $\mu(A_\varepsilon)=0$.
This lower bound away from zero lets us divide by $\varphi-\lambda$ outside a null set. Define $g:\Omega\to\mathbb C$ as follows: for $\omega\in\Omega\setminus A_\varepsilon$, set $g(\omega)=(\varphi(\omega)-\lambda)^{-1}$, and for $\omega\in A_\varepsilon$, set $g(\omega)=0$.
The function $g$ is measurable because it is obtained from the measurable function $\varphi-\lambda$ by inversion on the measurable set where the denominator is bounded away from zero, and by assigning the value $0$ on the remaining measurable set. On $\Omega\setminus A_\varepsilon$ we have $|\varphi(\omega)-\lambda|\ge\varepsilon$, hence
\begin{align*}
|g(\omega)|\le \varepsilon^{-1}.
\end{align*}
On $A_\varepsilon$ we have $g(\omega)=0$. Thus $g\in L^\infty(\Omega,\mathcal F,\mu)$, and multiplication by $g$ defines a bounded operator $M_g\in\mathcal L(H)$.
Now we verify that $M_g$ is the two-sided inverse of $M_\varphi-\lambda I$. For any $f\in H$, the identity
\begin{align*}
g(\varphi-\lambda)f=f
\end{align*}
holds outside the null set $A_\varepsilon$. Since elements of $L^2(\Omega,\mathcal F,\mu;\mathbb C)$ are equivalence classes modulo equality $\mu$-a.e., this gives
\begin{align*}
M_g(M_\varphi-\lambda I)f=f.
\end{align*}
The same pointwise computation also gives
\begin{align*}
(\varphi-\lambda)gf=f
\end{align*}
$\mu$-a.e., and therefore
\begin{align*}
(M_\varphi-\lambda I)M_g f=f.
\end{align*}
Since both identities hold for every $f\in H$, we have
\begin{align*}
M_g(M_\varphi-\lambda I_H)=I_H
\end{align*}
and
\begin{align*}
(M_\varphi-\lambda I_H)M_g=I_H.
\end{align*}
Thus $M_\varphi-\lambda I_H$ is boundedly two-sided invertible in $\mathcal L(H)$, so by the definition of the resolvent set $\lambda$ belongs to $\rho(M_\varphi)$. Therefore every spectral value of $M_\varphi$ must lie in $\operatorname{ess\,ran}(\varphi)$.
[/guided]
[/step]
[step:Construct approximate eigenvectors inside the essential range]
Fix $\lambda\in\operatorname{ess\,ran}(\varphi)$. For each positive integer $n\in\mathbb N$, define
\begin{align*}
A_n:=\{\omega\in\Omega:|\varphi(\omega)-\lambda|<n^{-1}\}.
\end{align*}
By the definition of essential range, $\mu(A_n)>0$ for every $n\in\mathbb N$. Since $(\Omega,\mathcal F,\mu)$ is semifinite, for every $n\in\mathbb N$ there exists $E_n\in\mathcal F$ such that
\begin{align*}
E_n\subseteq A_n
\end{align*}
and
\begin{align*}
0<\mu(E_n)<\infty.
\end{align*}
Define $u_n\in H$ by
\begin{align*}
u_n:=\mu(E_n)^{-1/2}\mathbb 1_{E_n}.
\end{align*}
Then
\begin{align*}
\|u_n\|_H^2=\mu(E_n)^{-1}\int_\Omega \mathbb 1_{E_n}\,d\mu(\omega)=1.
\end{align*}
Moreover,
\begin{align*}
\|(M_\varphi-\lambda I)u_n\|_H^2=\mu(E_n)^{-1}\int_{E_n}|\varphi(\omega)-\lambda|^2\,d\mu(\omega).
\end{align*}
Since $E_n\subseteq A_n$, we have $|\varphi(\omega)-\lambda|<n^{-1}$ for every $\omega\in E_n$, so
\begin{align*}
\|(M_\varphi-\lambda I)u_n\|_H^2\le n^{-2}.
\end{align*}
Thus
\begin{align*}
\|(M_\varphi-\lambda I)u_n\|_H\le n^{-1}.
\end{align*}
If $M_\varphi-\lambda I_H$ were boundedly two-sided invertible in $\mathcal L(H)$, then its inverse $B:=(M_\varphi-\lambda I_H)^{-1}\in\mathcal L(H)$ would satisfy, for every positive integer $n\in\mathbb N$,
\begin{align*}
1=\|u_n\|_H=\|B(M_\varphi-\lambda I_H)u_n\|_H\le \|B\|_{\mathcal L(H)}\|(M_\varphi-\lambda I_H)u_n\|_H\le \|B\|_{\mathcal L(H)}n^{-1}.
\end{align*}
Letting $n\to\infty$ gives a contradiction. Therefore $M_\varphi-\lambda I_H$ is not boundedly two-sided invertible in $\mathcal L(H)$, so by the definition of the spectrum $\lambda\in\sigma(M_\varphi)$. Hence
\begin{align*}
\operatorname{ess\,ran}(\varphi)\subseteq\sigma(M_\varphi).
\end{align*}
Together with the previous step, this proves
\begin{align*}
\sigma(M_\varphi)=\operatorname{ess\,ran}(\varphi).
\end{align*}
[/step]
[step:Identify the eigenvalues by solving the multiplication equation]
Fix $\lambda\in\mathbb C$, and define the level set
\begin{align*}
L_\lambda:=\{\omega\in\Omega:\varphi(\omega)=\lambda\}.
\end{align*}
First suppose $\lambda\in\sigma_p(M_\varphi)$. Then there exists $f\in H$ with $f\ne 0$ such that
\begin{align*}
M_\varphi f=\lambda f.
\end{align*}
Equivalently,
\begin{align*}
(\varphi-\lambda)f=0
\end{align*}
$\mu$-a.e. On $\Omega\setminus L_\lambda$ the scalar $\varphi(\omega)-\lambda$ is nonzero, so this equality implies $f=0$ $\mu$-a.e. on $\Omega\setminus L_\lambda$. If $\mu(L_\lambda)=0$, then $f=0$ $\mu$-a.e. on all of $\Omega$, contradicting $f\ne0$ in $H$. Therefore
\begin{align*}
\mu(L_\lambda)>0.
\end{align*}
Conversely, suppose $\mu(L_\lambda)>0$. Since $(\Omega,\mathcal F,\mu)$ is semifinite, there exists $E\in\mathcal F$ such that
\begin{align*}
E\subseteq L_\lambda
\end{align*}
and
\begin{align*}
0<\mu(E)<\infty.
\end{align*}
Define $u\in H$ by
\begin{align*}
u:=\mathbb 1_E.
\end{align*}
Then $u\ne0$ in $H$ because $\mu(E)>0$, and $u\in H$ because
\begin{align*}
\|u\|_H^2=\int_\Omega \mathbb 1_E\,d\mu(\omega)=\mu(E)<\infty.
\end{align*}
Since $E\subseteq L_\lambda$, we have $\varphi u=\lambda u$ $\mu$-a.e., and hence
\begin{align*}
M_\varphi u=\lambda u.
\end{align*}
Thus $\lambda\in\sigma_p(M_\varphi)$.
We have proved, for every $\lambda\in\mathbb C$,
\begin{align*}
\lambda\in\sigma_p(M_\varphi)\iff \mu(\{\omega\in\Omega:\varphi(\omega)=\lambda\})>0.
\end{align*}
[/step]
