Let \((X, \mathcal M, \mu)\) be a \(\sigma\)-finite measure space and let \(\mathcal E\subseteq \mathcal M\) be a collection containing \(\varnothing, X\). Then the following are equivalent:
For every \(E\in \mathcal M\) and every \(\varepsilon >0\) there exists a set \(G\in \mathcal E\) such that \[G\supseteq E\quad \text{ and } \quad \mu(G\setminus E)\le \varepsilon.\]
\(\mu\) is outer regular with respect to the family \(\mathcal E\), i.e. \[\mu(E)=\inf\{\mu(U): U\supseteq E \text{ and } U\in\mathcal E\} \ \text{ for any }\ E\in \mathcal M.\]
Proof. Suppose that (a) holds and \(\mu(E)=\infty\). Then for any set \(U\in\mathcal E\) such that \(U\supseteq E\) we have \(\mu(U)\ge \mu(E)=\infty\), thus \[\inf\{\mu(U): U\supseteq E \text{ and } U\in\mathcal E\}=\infty=\mu(E).\]
Suppose that (a) holds and \(\mu(E)<\infty\).
Let \[\alpha_E=\inf\{\mu(U): U\supseteq E \text{ and } U\in\mathcal E\}.\] we have to prove that \(\alpha_E=\mu(E)\) for any \(E\in \mathcal M\).
For any set \(U\in\mathcal E\) such that \(U\supseteq E\) we have \(\mu(E)\le \mu(U)\), yielding \(\mu(E)\le \alpha_E\).
By (a) for any \(\varepsilon>0\), we can find a set \(G\in\mathcal E\) such that \(G\supseteq E\) and \(\mu(G\setminus E)\le \varepsilon\). But \(\mu(E)<\infty\) thus \[\alpha_E\le \mu(G)\le \mu(E)+\varepsilon.\] Since \(\varepsilon>0\) is arbitrary we obtain \[\alpha_E\le \mu(E)\] as desired. $$\tag*{$\blacksquare$}$$
Suppose that (b) holds and \(\mu(E)<\infty\).
Using (b) for every \(\varepsilon>0\) there is a set \(G\in\mathcal E\) such that \(G\supseteq E\) and \(\mu(G) \leq \mu(E)+\varepsilon.\) But \(\mu(E)<\infty\) thus \(\mu(G\setminus E)\le \varepsilon\).
Suppose that (b) holds and \(\mu(E)=\infty\).
Since \(\mu\) is \(\sigma\)-finite thus \(X=\bigcup_{n\in \mathbb N}X_n\) for a sequence \((X_n)_{n\in \mathbb N}\subseteq \mathcal M\) of disjoint sets such that \(\mu(X_n)<\infty\). Setting \(E_n=E\cap X_n\) for \(n\in\mathbb N\) one sees that \((E_n)_{n\in \mathbb N}\subseteq \mathcal M\) and \(E=\bigcup_{n\in \mathbb N}E_n\) and \(\mu(E_n)<\infty\).
Applying the previous result to \(E_n\) for each \(n\in \mathbb N\) we obtain that for every \(\varepsilon>0\) there exists a set \(G_n\in\mathcal E\) such that \(G_n\supseteq E_n\) and \[\mu(G_n\setminus E_n)\le \varepsilon2^{-n}.\] Hence taking \(G=\bigcup_{n\in\mathbb N}G_n\) we have \[\mu(G\setminus E)\le \sum_{n\in\mathbb N}\mu(G_n\setminus E)\le \sum_{n\in\mathbb N}\mu(G_n\setminus E_n)\le \sum_{n\in\mathbb N}\varepsilon2^{-n}\le \varepsilon. \tag*{$\blacksquare$}\]
Let \((X, \mathcal M, \mu)\) be a \(\sigma\)-finite measure space and let \(\mathcal E\subseteq \mathcal M\) be a collection containing \(\varnothing, X\). Suppose that \(\mu\) is outer regular with respect to the family \(\mathcal E\), i.e. \[\mu(E)=\inf\{\mu(U): U\supseteq E \text{ and } U\in\mathcal E\} \ \text{ for any }\ E\in \mathcal M.\] Then for every \(E\in \mathcal M\) there exists a set \(G\in \mathcal E_{\delta}\) such that \(\mu(G\setminus E)=0\).
Proof. Let \(E\in \mathcal M\), then by the previous theorem for every \(n\in\mathbb N\) with \(\varepsilon=1/n\) we find a set \(U_n\in\mathcal E\) such that \(U_n\supseteq E\) and \(\mu(U_n\setminus E)\le 1/n\). Define \(G=\bigcap_{n\in\mathbb N}U_n\in\mathcal E_\delta\) and note that \[\mu(G\setminus E)\le \mu(U_n\setminus E)\le 1/n\] for every \(n\in\mathbb N\), which implies \(\mu(G\setminus E)=0\).$$\tag*{$\blacksquare$}$$
Let \((X, \mathcal M, \mu)\) be a \(\sigma\)-finite measure space and let \(\mathcal E\subseteq \mathcal M\) be a collection containing \(\varnothing, X\). Then the following are equivalent:
For every \(E\in \mathcal M\) and every \(\varepsilon >0\) there exists a set \(F\in\mathcal E\) such that \[ F\subseteq E \quad \text{ and } \quad \mu(E\setminus F)\le \varepsilon.\]
\(\mu\) is inner regular with respect to the family \(\mathcal E\), i.e. \[\mu(E)=\sup\{\mu(C): C\subseteq E \text{ and } C\in\mathcal E\} \ \text{ for any }\ E\in \mathcal M.\]
Proof. Suppose that (a) holds.
Define \[\alpha_E=\sup\{\mu(C): C\subseteq E \text{ and } C\in\mathcal E\}.\] we have to prove that \(\alpha_E=\mu(E)\) for any \(E\in \mathcal M\).
For any set \(C\in\mathcal E\) such that \(C\subseteq E\) we have \(\mu(C)\le \mu(E)\), yielding \(\alpha_E\le \mu(E)\).
By (a) for any \(\varepsilon>0\), we can find a set \(F\in\mathcal E\) such that \(F\subseteq E\) and \(\mu(E\setminus F)\le \varepsilon\). Thus \[\mu(E)=\mu(F) + \mu(E\setminus F)\le \mu(F)+\varepsilon\le \alpha_E+\varepsilon.\] Since \(\varepsilon>0\) is arbitrary we obtain \(\mu(E)\le \alpha_E\) as desired. $$\tag*{$\blacksquare$}$$
Suppose that (b) holds and \(\mu(E)<\infty\).
Using (b) for every \(\varepsilon>0\) there is a set \(C\in\mathcal E\) such that \(C\subseteq E\) and \(\mu(E) \leq \mu(C)+\varepsilon.\) But \(\mu(C)<\infty\) thus \(\mu(E\setminus C)\le \varepsilon\).
Suppose that (b) holds and \(\mu(E)=\infty\).
Suppose that \(\mu(E)=\infty\) and \(E=\bigcup_{j\in\mathbb N}E_j\), where \(E_j\in\mathcal M\) and \(E_j\subseteq E_{j+1}\) and \(\mu(E_j)<\infty\) for all \(j\in\mathbb N\). Thus for any \(N\in\mathbb N\) there is \(j\in\mathbb N\) such that \(\mu(E_j)>N\). By the preceding argument we find a set \(C\in\mathcal E\) such that \(C\subseteq E_j\) with \(\mu(C)>N\), yielding \(\alpha_E=\infty\).$$\tag*{$\blacksquare$}$$
Let \((X, \mathcal M, \mu)\) be a \(\sigma\)-finite measure space and let \(\mathcal E\subseteq \mathcal M\) be a collection containing \(\varnothing, X\). Suppose that \(\mu\) is inner regular with respect to the family \(\mathcal E\), i.e. \[\mu(E)=\sup\{\mu(C): C\subseteq E \text{ and } C\in\mathcal E\} \ \text{ for any }\ E\in \mathcal M.\] Then for every \(E\in \mathcal B_X\) there exists a set \(F\in \mathcal E_{\sigma}\) such that \(\mu(E\setminus F)=0\).
Proof. Let \(E\in \mathcal M\), then by the previous theorem for every \(n\in\mathbb N\) with \(\varepsilon=1/n\) we find a set \(C_n\in\mathcal E\) such that \(C_n\subseteq E\) and \(\mu(E\setminus C_n)\le 1/n\). Define \(F=\bigcup_{n\in\mathbb N}C_n\in\mathcal E_\sigma\) and note that \[\mu(E\setminus F)\le \mu(E\setminus C_n)\le 1/n\] for every \(n\in\mathbb N\), which implies \(\mu(E\setminus F)=0\).$$\tag*{$\blacksquare$}$$
Notations
Unless otherwise stated, \(X\) is a topological space.
\(\mathcal U_X={\rm Open}(X)\) is the is the collection of all open sets in \(X\).
\(\mathcal F_X={\rm Closed}(X)\) is the is the collection of all closed sets in \(X\).
\(\mathcal K_X={\rm Compact}(X)\) is the collection of all compact sets in \(X\),
\(\mathcal B_X={\rm Bor}(X)=\sigma(\mathcal U_X)=\sigma(\mathcal F_X)\) is the Borel \(\sigma\)-algebra on \(X\).
Borel measures A measure \(\mu\) defined on a Borel \(\sigma\)-algebra \(\mathcal B_X\) is called a Borel measure. A pair \((X, \mathcal B_X)\) is called a Borel space, whereas a triple \((X, \mathcal B_X, \mu)\) is called a Borel measure space.
Let \((X, \mathcal B_X, \mu)\) be a finite Borel measure space such that every closed set in \(X\) is a \(G_{\delta}\) set. Then for every \(E\in \mathcal B_X\) and every \(\varepsilon >0\) there are two sets: an open set \(G\supseteq E\) and a closed set \(F\subseteq E\) such that \[\mu(E\setminus F)\le \varepsilon \qquad \text{ and } \qquad \mu(G\setminus E)\le \varepsilon.\]
Remark. If \((X, d)\) is a metric set then every closed set in \(X\) is a \(G_{\delta}\) set in \(X\).
Proof of the Remark. Let \(F\) be a closed set in \(X\) and observe that \[F=\bigcap_{n\in\mathbb N}\Big\{x\in X: d(x, F)<\frac{1}{n}\Big\},\] as desired.$$\tag*{$\blacksquare$}$$
Proof. Let \(\mathcal M\) be the set of all Borel sets \(E\in \mathcal B_X\) with the property that for every \(\varepsilon >0\) there are two sets: an open set \(G\supseteq E\) and a closed set \(F\subseteq E\) such that \[\mu(E\setminus F)\le \varepsilon \qquad \text{ and } \qquad \mu(G\setminus E)\le \varepsilon.\] We show that \(\mathcal M\) is a \(\sigma\)-algebra containing all open subsets of \(X\). This in turn will imply that \(\mathcal M=\mathcal B_X\), and the proof will be finished.
\(\mathcal M\) is closed under complements. Indeed, if \(E\in\mathcal M\) then for every \(\varepsilon >0\) there are two sets: closed \(F\) and open \(G\) such that \[F\subseteq E\subseteq G\quad \Longleftrightarrow\quad G^c\subseteq E^c\subseteq F^c \qquad\text{ and }\] \[\mu(E^c\setminus G^c)=\mu(G\setminus E)\le \varepsilon \qquad \text{ and } \qquad \mu(F^c\setminus E^c)=\mu(E\setminus F)\le \varepsilon,\]
and we are done since \(G^c\) is closed and \(F^c\) is open.
\(\mathcal M\) is closed under countable unions. Let \(E_1, E_2,\ldots\in\mathcal M\) and let \(E=\bigcup_{n\in \mathbb N}E_n\). For every \(\varepsilon >0\) and every \(n\in\mathbb N\) there are closed \(F_n\) and open \(G_n\) sets such that \(F_n\subseteq E_n\subseteq G_n\) and \[\mu(G_n\setminus E_n)\le \varepsilon2^{-n} \qquad \text{ and } \qquad \mu(E_n\setminus F_n)\le \varepsilon2^{-{n-1}}.\] Define an open set \(G=\bigcup_{n\in \mathbb N}G_n\) and note that \(E\subseteq G\) and \[\mu(G\setminus E)\le \sum_{n\in\mathbb N}\mu(G_n\setminus E)\le \sum_{n\in\mathbb N}\mu(G_n\setminus E_n)\le \sum_{n\in\mathbb N}\varepsilon2^{-n}\le \varepsilon.\]
Consider \(F_0=\bigcup_{n\in \mathbb N}F_n\) and note that \(F_0\subseteq E\). By the continuity and finiteness of \(\mu\) there is \(m\in\mathbb N\) such that \[\mu\bigg(F_0\setminus\bigcup_{n=1}^mF_n\bigg)\le \varepsilon/2.\]
Define a closed set \(F=\bigcup_{n=1}^mF_n\) and note that \(F\subseteq F_0\subseteq E\) and \(E\setminus F=E\cap F^c\cap F_0^c\cup E\cap F^c\cap F_0\subseteq E\setminus F_0\cup F_0\setminus F\). \[\mu(E\setminus F)\le \mu(E\setminus F_0)+\mu(F_0\setminus F)\le \varepsilon,\] since \(\mu(F_0\setminus F)=\mu\big(F_0\setminus\bigcup_{n=1}^mF_n\big)\le \varepsilon/2\), and \[\mu(E\setminus F_0)\le \sum_{n\in\mathbb N}\mu(E_n\setminus F_0)\le \sum_{n\in\mathbb N}\mu(E_n\setminus F_n)\le \sum_{n\in\mathbb N}\varepsilon2^{-n-1}\le \varepsilon/2.\]
\(\mathcal M\) contains open sets. Let \(O\in \mathcal U_X\) be an open set. Fix \(\varepsilon >0\). Taking \(G=O\) we see that \(\mu(G\setminus O)=0< \varepsilon\). Now since every closed set in \(X\) is a \(G_{\delta}\) set then \(O=\bigcup_{n\in \mathbb N}F_n\) for some sequence \((F_n)_{n\in \mathbb N}\) of closed sets. By the continuity and finiteness of \(\mu\) there is \(m\in\mathbb N\) such that \(\mu\big(O\setminus\bigcup_{n=1}^mF_n\big)\le \varepsilon\). Taking a closed set \(F=\bigcup_{n=1}^mF_n\) we obtain the desired claim. $$\tag*{$\blacksquare$}$$
Let \((X, \mathcal B_X, \mu)\) be a Borel measure space such that every closed set in \(X\) is a \(G_{\delta}\) set. Assume that \(X=\bigcup_{n\in \mathbb N} V_n\) for some open sets \(V_n\) such that \(\mu(V_n)<\infty\) for every \(n\in\mathbb N\). Then for every \(E\in \mathcal B_X\) and every \(\varepsilon >0\) there are two sets: an open set \(G\supseteq E\) and a closed set \(F\subseteq E\) such that \[\mu(E\setminus F)\le \varepsilon \qquad \text{ and } \qquad \mu(G\setminus E)\le \varepsilon.\]
Remark. Suppose that \((X, \mathcal B_X, \mu)\) is a Borel measure space that is finite on every compact set. If additionally \(X\) is a locally compact Hausdorff space which is \(\sigma\)-compact (i.e. \(X\) is a countable union of compact sets) then there exists a sequence \((U_n)_{n\in\mathbb N}\) of open sets with compact closures such that \({\rm cl}(U_n)\subseteq U_{n+1}\) for all \(n\in \mathbb N\) and \(X=\bigcup_{n\in \mathbb N} U_n\). Then we immediately obtain that \[\mu(U_n)<\infty.\]
Proof. We shall reduce the matters to the previous theorem. Decompose \(X=\bigcup_{n\in \mathbb N} V_n\) for some open sets \(V_n\) such that \(\mu(V_n)<\infty\) for all \(n\in\mathbb N\).
Let \(E\in \mathcal B_X\), fix \(\varepsilon>0\) and define \(E_n=E\cap V_n\) for every \(n\in\mathbb N\).
For each \(n\in\mathbb N\) consider a \(\sigma\)-algebra \(\mathcal B_X\upharpoonright{V_n}=\{E\cap V_n: E\in \mathcal B_X\}\) with a measure \(\mu_{\upharpoonright{V_n}}\), which is a restriction of \(\mu\) to \(\mathcal B_X\upharpoonright{V_n}\). Then \((V_n, \mathcal B_X\upharpoonright{V_n}, \mu_{\upharpoonright{V_n}})\) is a finite Borel measure space with the property that every closed set in \(V_n\) is a \(G_{\delta}\) set. Therefore the conclusion of the previous theorem is true for this Borel measure space.
For each \(n\in\mathbb N\) we can find open sets \(G_n\in \mathcal B_X\upharpoonright{V_n}\), which are also open in \(X\) such that \(G_n\supseteq E_n\) and \(\mu(G_n\setminus E_n)\le \varepsilon 2^{-n}\). Taking \(G=\bigcup_{n\in \mathbb N} G_n\) we see that \[\mu(G\setminus E)\le \sum_{n\in\mathbb N}\mu(G_n\setminus E)\le \sum_{n\in\mathbb N}\mu(G_n\setminus E_n)\le \sum_{n\in\mathbb N}\varepsilon2^{-n}\le \varepsilon.\]
Applying the same reasoning with \(E^c\) in place of \(E\) we obtain \(\mu(E\setminus F)=\mu(E\setminus G^c)=\mu(G\setminus E^c)\le \varepsilon\) with a closed \(F=G^c\).$$\tag*{$\blacksquare$}$$
We say that \(\mu\) is outer regular for \(E\in \mathcal B_X\) if \[\mu(E)=\inf\{\mu(U): U\supseteq E\text{ and } U\in\mathcal U_X\}.\]
We say that \(\mu\) is inner regular for \(E\in \mathcal B_X\) if \[\mu(E)=\sup\{\mu(K): K\subseteq E\text{ and } K\in\mathcal K_X\}.\]
We say that \(\mu\) is regular for \(E \in \mathcal B_X\) if \(\mu\) is both outer and inner regular for \(E \in \mathcal B_X\).
Let \(\mathcal V\subseteq \mathcal B_X\) be an arbitrary collection. We say that \(\mu\) is outer regular for \(\mathcal V\), inner regular for \(\mathcal V\), or regular for \(\mathcal V\) according as \(\mu\) is outer regular, inner regular, or regular for every \(E\in \mathcal V\).
Let \((X, \mathcal B_X, \mu)\) be a \(\sigma\)-finite Borel measure space. Suppose that
\(\mu\) is outer regular for \(E\in \mathcal B_X\), i.e. \[\mu(E)=\inf\{\mu(U): U\supseteq E \text{ and } U\in\mathcal U_X\}.\]
\(\mu\) is \(\mu\) is inner regular for \(E\in \mathcal B_X\), i.e. \[\mu(E)=\sup\{\mu(K): K\subseteq E\text{ and } K\in\mathcal K_X\}.\]
Then
For every \(E\in \mathcal B_X\) and every \(\varepsilon >0\) there exists a closed set \(C\) and an open set \(U\) such that \(C\subseteq E\subseteq U\) and \(\mu(U\setminus C)\le \varepsilon.\)
For every \(E\in \mathcal B_X\) there exists an \(F_\sigma\) set \(F\) and a \(G_\delta\) set \(G\) such that \(F\subseteq E\subseteq G\) and \(\mu(G\setminus F)=0.\)
Outer regular for \(\mathcal B_X\): \[\mu(E)=\inf\{\mu(U): U\supseteq E\text{ and } U\in\mathcal U_X\} \quad \text{ for every } \quad E \in \mathcal B_X.\]
Inner regular for \(\mathcal U_X\): \[\mu(U)=\sup\{\mu(K): K\subseteq U\text{ and } K\in\mathcal K_X\} \quad \text{ for every } \quad U \in \mathcal U_X.\]
Finite on \(\mathcal K_X\): \[\mu(K)<\infty \quad \text{ for every } \quad K \in \mathcal K_X.\]
Let \(X\) be a Hausdorff space. Any \(\sigma\)-finite Radon measure \(\mu\) on a Borel space \((X, \mathcal B_X)\) is regular for \(\mathcal B_X\). In particular, every finite Radon measure is regular for \(\mathcal B_X\).
Proof. We have to show that \(\mu(E)=\alpha_E\), where \[\alpha_E=\sup\{\mu(K): K\subseteq E\text{ and } K\in\mathcal K_X\}.\] for every \(E\in\mathcal B_X\). Suppose that \(\mu(E)<\infty\).
Note that \(\alpha_E\le \mu(E)\) since \(\mu(K)\le \mu(E)\) for every compact \(K\subseteq E\).
It suffices to prove that \(\alpha_E \ge \mu(E)\). Fix \(\varepsilon>0\).
By the outer regularity for \(\mathcal B_X\) we find an open set \(U\supseteq E\) such that \[\mu(U)\le \mu(E)+\varepsilon/2.\]
By the inner regularity for \(\mathcal U_X\) we find a compact set \(F\subseteq U\) such that \[\mu(F)\ge \mu(U)-\varepsilon/2.\]
By the outer regularity for \(\mathcal B_X\) again, since \(\mu(U\setminus E)\le \varepsilon/2\) we can also choose and open \(V\supseteq U\setminus E\) such that \(\mu(V)\le \varepsilon/2\).
Consider a compact set \(K=F\setminus V\) and note that \[K=F\cap V^c\subseteq F\cap(U^c\cup E) \subseteq E,\] since \(V^c\subseteq (U\cap E^c)^c=U^c\cup E\) and \(F\cap U^c=\varnothing\).
Furthermore, since \(\mu(F)\ge \mu(U)-\varepsilon/2\) and \(E\subseteq U\) we obtain \[\alpha_E \ge \mu(K)=\mu(F)-\mu(F\cap V)\ge \mu(U)-\varepsilon/2-\mu(V)\ge \mu(E)-\varepsilon,\] yielding \(\alpha_E \ge \mu(E)\) as desired.
Suppose that \(\mu(E)=\infty\) and \(E=\bigcup_{j\in\mathbb N}E_j\), where \(E_j\in\mathcal B_X\) and \(E_j\subseteq E_{j+1}\) and \(\mu(E_j)<\infty\) for all \(j\in\mathbb N\). Thus for any \(N\in\mathbb N\) there is \(j\in\mathbb N\) such that \(\mu(E_j)>N\) and by the preceding argument we find a compact \(K\subseteq E_j\) with \(\mu(K)>N\), which ensures that \(\alpha_E=\infty\).$$\tag*{$\blacksquare$}$$
Suppose that \(X\) is a locally compact Hausdorff space in which every open set is \(\sigma\)-compact (i.e. is a countable union of compact sets). Then every Borel measure on \((X, \mathcal B_X)\) that is finite on compact sets is Radon.
Proof. A few comments are in order.
Note that every closed set is a \(G_\delta\) set, since every open set is \(\sigma\)-compact in \(X\).
Since \(X\) is a locally compact Hausdorff space which is \(\sigma\)-compact, then there exists a sequence \((U_n)_{n\in\mathbb N}\) of open sets with compact closures such that \({\rm cl}(U_n)\subseteq U_{n+1}\) for all \(n\in \mathbb N\) and \(X=\bigcup_{n\in \mathbb N} U_n\). Then we immediately obtain that \[\mu(U_n)<\infty.\]
Therefore \(\mu\) is outer regular for \(\mathcal B_X\): \[\mu(E)=\inf\{\mu(U): U\supseteq E\text{ and } U\in\mathcal U_X\} \quad \text{ for every } \quad E \in \mathcal B_X.\]
We prove that \(\mu\) is inner regular for \(\mathcal U_X\): \[\mu(U)=\sup\{\mu(K): K\subseteq U\text{ and } K\in\mathcal K_X\} \quad \text{ for every } \quad U \in \mathcal U_X.\] Indeed, every open set \(U\) is \(\sigma\)-compact. Let \(U=\bigcup_{n\in \mathbb N}K_n\) for some increasing sequence \((K_n)_{n\in \mathbb N}\) of compact sets. If \(\mu(U)<\infty\) then by the continuity of the measure \(\mu\) for any \(\varepsilon>0\) we find \(n\in \mathbb N\) such that \(\mu(K_n)\ge \mu(U)-\varepsilon\), which impies \(\mu(U\setminus K_n)\le \varepsilon\) as desired. If \(\mu(U)=\infty\) then for any \(M>0\) there exists \(n\in \mathbb N\) such that \(\mu(K_n)\ge M\), which also gives the desired claim.
Invoking the last theorem we know that in \(\sigma\)-finite measure spaces inner regularity for \(\mathcal U_X\) becomes regularity for \(\mathcal B_X\).$$\tag*{$\blacksquare$}$$