[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 $E(V_n)=0$.[/step]

[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: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]