[proofplan]
We identify $K\setminus S$ as the union of all open sets on which the projection-valued measure vanishes. Since $K$ is a compact [metric space](/page/Metric%20Space), it is second countable; we prove this directly and use it to choose a countable subcover of $K\setminus S$ by such null open sets. Countable additivity of the scalar measures associated to $E$ then shows that this countable union is $E$-null. The final assertion is the minimality property of the support: any [closed set](/page/Closed%20Set) whose complement is $E$-null must contain every point whose neighbourhoods all have nonzero projection.
[/proofplan]
[step:Describe the complement of the support by null open neighbourhoods]
Let $\mathcal{N}$ denote the family of open subsets $U\subset K$ such that $E(U)=0$. By the definition of $S=\operatorname{supp}(E)$, a point $x\in K$ lies outside $S$ exactly when there exists an open neighbourhood $U\subset K$ of $x$ such that $E(U)=0$. Hence
\begin{align*}
K\setminus S=\bigcup_{U\in\mathcal{N}} U.
\end{align*}
[/step]
[step:Extract a countable null open cover of $K\setminus S$]
We first record the needed countability property. Since $K$ is compact and metric, for each $m\in\mathbb{N}$ there exists a finite set $A_m\subset K$ such that
\begin{align*}
K=\bigcup_{a\in A_m} B(a,1/m).
\end{align*}
Let
\begin{align*}
A=\bigcup_{m=1}^{\infty} A_m.
\end{align*}
Then $A$ is countable and dense in $K$. The family
\begin{align*}
\mathcal{B}_0=\{B(a,r): a\in A,\ r\in\mathbb{Q},\ r>0\}
\end{align*}
is countable and is a base for the metric topology of $K$.
For each $x\in K\setminus S$, choose $U_x\in\mathcal{N}$ with $x\in U_x$. Because $\mathcal{B}_0$ is a base, there exists $B_x\in\mathcal{B}_0$ such that
\begin{align*}
x\in B_x\subset U_x.
\end{align*}
Since $B_x\subset U_x$ and $E(U_x)=0$, the projection-valued measure identity
\begin{align*}
E(B_x)=E(B_x)E(U_x)
\end{align*}
gives $E(B_x)=0$. Thus each selected $B_x$ belongs to $\mathcal{N}$. Because the selected sets all lie in the countable family $\mathcal{B}_0$, their distinct values may be listed as $(V_n)_{n=1}^{\infty}$ after omitting repetitions and empty surplus terms. Therefore
\begin{align*}
K\setminus S=\bigcup_{n=1}^{\infty} V_n,
\end{align*}
where each $V_n\subset K$ is open and satisfies $E(V_n)=0$.
[guided]
The purpose of this step is to replace a possibly uncountable union of null open sets by a countable one. Countable additivity only directly controls countable unions, so the metric compactness hypothesis is used exactly here.
We first prove that $K$ has a countable base. Since $K$ is compact, the open cover
\begin{align*}
\{B(x,1/m):x\in K\}
\end{align*}
has a finite subcover for every $m\in\mathbb{N}$. Choose a finite set $A_m\subset K$ such that
\begin{align*}
K=\bigcup_{a\in A_m} B(a,1/m).
\end{align*}
Define
\begin{align*}
A=\bigcup_{m=1}^{\infty} A_m.
\end{align*}
This set is countable because it is a countable union of finite sets. It is dense in $K$: if $x\in K$ and $\varepsilon>0$, choose $m\in\mathbb{N}$ with $1/m<\varepsilon$; then some $a\in A_m$ satisfies $x\in B(a,1/m)$, hence $d(x,a)<\varepsilon$.
Now define
\begin{align*}
\mathcal{B}_0=\{B(a,r): a\in A,\ r\in\mathbb{Q},\ r>0\}.
\end{align*}
This family is countable because $A$ and the positive rationals are countable. It is a base: if $O\subset K$ is open and $x\in O$, choose $\varepsilon>0$ such that $B(x,\varepsilon)\subset O$. Choose $a\in A$ with $d(x,a)<\varepsilon/4$ and choose $r\in\mathbb{Q}$ with
\begin{align*}
d(x,a)<r<\varepsilon/2.
\end{align*}
Then $x\in B(a,r)$. Also, if $y\in B(a,r)$, the triangle inequality gives
\begin{align*}
d(y,x)\le d(y,a)+d(a,x)<r+\varepsilon/4<\varepsilon,
\end{align*}
so $B(a,r)\subset B(x,\varepsilon)\subset O$.
For each $x\in K\setminus S$, the definition of $S$ gives an open neighbourhood $U_x\subset K$ of $x$ with $E(U_x)=0$. Since $\mathcal{B}_0$ is a base, choose $B_x\in\mathcal{B}_0$ such that
\begin{align*}
x\in B_x\subset U_x.
\end{align*}
Because $B_x\subset U_x$, the projection-valued measure multiplication rule gives
\begin{align*}
E(B_x)=E(B_x\cap U_x)=E(B_x)E(U_x).
\end{align*}
Since $E(U_x)=0$, it follows that $E(B_x)=0$. Thus every point of $K\setminus S$ is covered by a member of the countable base which is also $E$-null. Listing the distinct chosen base elements gives a sequence $(V_n)_{n=1}^{\infty}$ of open sets such that
\begin{align*}
K\setminus S=\bigcup_{n=1}^{\infty} V_n
\end{align*}
and $E(V_n)=0$ for every $n\in\mathbb{N}$.
[/guided]
[/step]
[step:Use countable additivity to prove that $K\setminus S$ is null]
Let
\begin{align*}
V=\bigcup_{n=1}^{\infty} V_n.
\end{align*}
Then $V=K\setminus S$. For each $x\in H$, define the scalar measure $\mu_x:\mathcal{B}(K)\to[0,\infty)$ by
\begin{align*}
\mu_x(\Delta)=(E(\Delta)x,x)_H.
\end{align*}
By [Countable Additivity in Matrix Coefficients][citetheorem:8404], $\mu_x$ is a finite positive measure. Since $E(V_n)=0$, we have
\begin{align*}
\mu_x(V_n)=0
\end{align*}
for every $n\in\mathbb{N}$. [Countable subadditivity](/theorems/1108) of the positive measure $\mu_x$ gives
\begin{align*}
\mu_x(V)\le \sum_{n=1}^{\infty}\mu_x(V_n)=0.
\end{align*}
Thus
\begin{align*}
(E(V)x,x)_H=0
\end{align*}
for every $x\in H$. Since $E(V)$ is an [orthogonal projection](/theorems/437), this implies $E(V)=0$. Therefore
\begin{align*}
E(K\setminus S)=0.
\end{align*}
[/step]
[step:Prove the minimality property of the support]
Let $F\subset K$ be closed and suppose that
\begin{align*}
E(K\setminus F)=0.
\end{align*}
Take $x\in K\setminus F$. Since $F$ is closed in $K$, the set $K\setminus F$ is open in $K$ and is an open neighbourhood of $x$. By hypothesis this neighbourhood is $E$-null. Hence $x\notin S$ by the definition of $S$. We have proved
\begin{align*}
K\setminus F\subset K\setminus S.
\end{align*}
Taking complements in $K$ gives
\begin{align*}
S\subset F.
\end{align*}
This proves the asserted characterisation of the support.
[/step]
