Definition. Let \({\rm v}_d:J^d\to[0, \infty]\) be the volume function on \(\mathbb R^d\). The \(d\)-dimensional Lebesgue outer measure is defined for any \(E \subseteq \mathbb{R}^d\) by setting \[\begin{align*} \lambda_d^{*}(E)&=\inf\Big\{\sum_{n \in \mathbb{N}}{\rm v}_d(E_n): (E_n)_{n \in \mathbb{N}}\subseteq J_{co}^d\quad \text{ and } \quad E \subseteq \bigcup_{n \in \mathbb{N}}E_n\Big\}. \end{align*}\]
The \(d\)-dimensional Lebesgue measure \(\lambda_d\) is the restriction of \(\lambda_d^*\) to the Carathéodory \(\sigma\)-algebra \(\mathcal{L}(\mathbb R^d)=\mathcal{M}(\lambda_d^{*})\) of \(\lambda_d^{*}\)-measurable sets.
\(\mathcal{L}(\mathbb R^d)=\mathcal{M}(\lambda_d^{*})\) will be called the Lebesgue \(\sigma\)-algebra.
The members of \(\mathcal{L}(\mathbb R^d)\) are the Lebesgue measurable sets.
By Carathéodory’s theorem \((\mathbb R^d, \mathcal{L}(\mathbb R^d), \lambda_d)\) is a complete measure space. For any rectangle \(E\in J^d=J_o^d \cup J_{oc}^d \cup J_{co}^d \cup J_c^d\), we have \(\lambda_d(E)={\rm v}_d(E)\) and \(\lambda_d\) is \(\sigma\)-finite on \(\mathbb R^d\), and \({\rm Bor}(\mathbb R^d)\subseteq \mathcal{L}(\mathbb R^d)\).
If \(d=1\) we shall abbreviate \(\lambda_d^*\) to \(\lambda^*\) and \(\lambda_d\) to \(\lambda\).
Theorem. For \(E \subseteq \mathbb{R}^d\) the following conditions are equivalent:
\(E \in \mathcal L(\mathbb{R}^d)\);
For every \(\varepsilon>0\) there exists an open set \(O \supseteq E\) so that \(\lambda_d^{*}(O \setminus E) \leq \varepsilon\);
There exists a \(G_{\delta}\)-set \(G \supseteq E\) with \(\lambda_d^{*}(G \setminus E)=0\);
For every \(\varepsilon>0\) there exists a closed set \(C \subseteq E\) with \(\lambda_d^*(E \setminus C) \leq \varepsilon\);
There exists an \(F_{\sigma}\) set \(F \subseteq E\) with \(\lambda_d^*(E \setminus F)=0\).
Moreover, if \(E \in \mathcal L(\mathbb{R}^d)\), then one has \[\begin{align*} \lambda_d(E)&=\inf\{\lambda_d(U): U\supseteq E \text{ and } U \text{ is open}\}\\ &=\sup\{\lambda_d(F): F\subseteq E \text{ and } F \text{ is closed}\}\\ &=\sup\{\lambda_d(K): K\subseteq E \text{ and } K \text{ is compact}\}. \end{align*}\]
Step 1. We show that \(\lambda_d^{*}(E)=\inf\{\lambda_d(U): U \supseteq E \text{ and } U \text{ is open}\}\) for every \(E \subseteq \mathbb R^d\). Let \(\alpha(E)=\inf\{\lambda_d(U): U \supseteq E \text{ and } U \text{ is open}\}\).
By monotonicity \(\lambda_d^{*}(E) \leq \lambda_d(U)\) for every open \(U \supseteq E\), so \(\lambda_d^{*}(E)\) is not larger than the infimum, i.e. \(\lambda_d^{*}(E) \leq \alpha(E)\).
For the converse we may assume that \(\lambda_d^{*}(E)<\infty\), since otherwise there is nothing to do. Let \(\varepsilon>0\) and choose \((I_n)_{n \in \mathbb N} \subseteq J_{co}^d\) such that \[E \subseteq \bigcup_{n \in \mathbb N}I_n \quad \text{ and } \quad \sum_{n \in \mathbb N}{\rm v}_d(I_n) \leq \lambda_d^{*}(E)+\frac{\varepsilon}{2}.\]
If \(I_n=[a_n, b_n)\) we may enlarge it slightly to an open rectangle \(U_n \supseteq I_n\) with \({\rm v}_d(U_n) \leq {\rm v}_d(I_n)+\varepsilon 2^{-n-1}\).
Then \(U=\bigcup_{n \in \mathbb N}U_n\) is open, \(E \subseteq U\) and, by subadditivity, \[\alpha(E)\le \lambda_d(U) \leq \sum_{n \in \mathbb N}{\rm v}_d(U_n) \leq \sum_{n \in \mathbb N}{\rm v}_d(I_n)+\frac{\varepsilon}{2} \leq \lambda_d^{*}(E)+\varepsilon.\]
Step 2. We show that (i) \(\Longrightarrow\) (ii). Let \(E \in \mathcal L(\mathbb R^d)\) and \(\varepsilon>0\).
Assume first that \(\lambda_d(E)<\infty\). By Step 1 there is an open set \(O \supseteq E\) with \(\lambda_d(O) \leq \lambda_d(E)+\varepsilon\). Since \(E\) is measurable and \(\lambda_d(E)<\infty\), \[\lambda_d^{*}(O \setminus E)=\lambda_d(O)-\lambda_d(E) \leq \varepsilon.\]
The assumption \(\lambda_d(E)<\infty\) is essential here, since otherwise we cannot subtract \(\lambda_d(E)\).
In the general case put \(E_n=E \cap (-n, n)^d\) for \(n \in \mathbb N\). Then we have \(\lambda_d(E_n)<\infty\), so by the previous case there are open sets \(O_n \supseteq E_n\) with \(\lambda_d^{*}(O_n \setminus E_n) \leq \varepsilon 2^{-n}\).
Since \(E=\bigcup_{n \in \mathbb N}E_n\), the set \(O=\bigcup_{n \in \mathbb N}O_n\) is open, \(E \subseteq O\) and \(O \setminus E \subseteq \bigcup_{n \in \mathbb N}(O_n \setminus E_n)\), hence \[\lambda_d^{*}(O \setminus E) \leq \sum_{n \in \mathbb N}\lambda_d^{*}(O_n \setminus E_n) \leq \sum_{n \in \mathbb N}\varepsilon 2^{-n}=\varepsilon.\]
Step 3. We show that (ii) \(\Longrightarrow\) (iii) \(\Longrightarrow\) (i).
Assume (ii). For every \(j \in \mathbb N\) choose an open set \(O_j \supseteq E\) such that \(\lambda_d^{*}(O_j \setminus E) \leq \frac{1}{j}\) and put \[G=\bigcap_{j \in \mathbb N}O_j.\]
Then \(G\) is a \(G_{\delta}\)-set, \(E \subseteq G\) and \(G \setminus E \subseteq O_j \setminus E\) for every \(j \in \mathbb N\), so by monotonicity \[\lambda_d^{*}(G \setminus E) \leq \inf_{j \in \mathbb N}\frac{1}{j}=0,\] which is (iii).
Assume (iii) and let \(G \supseteq E\) be a \(G_{\delta}\)-set with \(\lambda_d^{*}(G \setminus E)=0\).
Then \(G \in {\rm Bor}(\mathbb R^d) \subseteq \mathcal L(\mathbb R^d)\), and \(G \setminus E \in \mathcal L(\mathbb R^d)\) because \(\lambda_d\) is a complete measure and \(\lambda_d^{*}(G \setminus E)=0\). Consequently \[E=G \setminus (G \setminus E) \in \mathcal L(\mathbb R^d),\] which is (i).
Step 4. We show that (i) \(\Longleftrightarrow\) (iv) and (iii) \(\Longleftrightarrow\) (v) by passing to complements.
If \(C \subseteq E\) then \(O=C^c\) satisfies \(O \supseteq E^c\) and \[O \setminus E^c=O \cap E=E \setminus C,\] and \(C\) is closed if and only if \(O\) is open.
Hence \(E\) satisfies (iv) if and only if \(E^c\) satisfies (ii). Since \(E \in \mathcal L(\mathbb R^d)\) if and only if \(E^c \in \mathcal L(\mathbb R^d)\), the equivalence (i) \(\Longleftrightarrow\) (iv) follows from (i) \(\Longleftrightarrow\) (ii) applied to \(E^c\).
Similarly, \(F \subseteq E\) is an \(F_{\sigma}\)-set if and only if \(F^c \supseteq E^c\) is a \(G_{\delta}\)-set, and \(F^c \setminus E^c=E \setminus F\), so \(E\) satisfies (v) if and only if \(E^c\) satisfies (iii).
Step 5. We prove the first two equalities. Let \(E \in \mathcal L(\mathbb R^d)\).
Let \(\alpha(E)=\inf\{\lambda_d(U): U \supseteq E \text{ and } U \text{ is open}\}\). By monotonicity \(\lambda_d(E) \leq \alpha(E)\), and the reverse inequality is precisely Step 1.
Let \(\beta(E)=\sup\{\lambda_d(C): C \subseteq E \text{ and } C \text{ is closed}\}\). By monotonicity we also have \(\beta(E)\leq \lambda_d(E)\).
If \(\lambda_d(E)<\infty\), then by (iv) for every \(\varepsilon>0\) there is a closed set \(C \subseteq E\) with \(\lambda_d(E \setminus C) \leq \varepsilon\), whence \[\beta(E)\ge \lambda_d(C)=\lambda_d(E)-\lambda_d(E \setminus C) \geq \lambda_d(E)-\varepsilon.\]
If \(\lambda_d(E)=\infty\), then by (iv) there is a closed set \(C \subseteq E\) with \(\lambda_d(E \setminus C) \leq 1\), and since \(\lambda_d(E) \leq \lambda_d(C)+\lambda_d(E \setminus C)\) we obtain \(\lambda_d(C)=\infty\), which also gives \(\beta(E)\ge\lambda_d(E)\).
In both cases we have \(\beta(E)=\lambda_d(E)\).
Step 6. We prove the third equality.
Set \[\begin{align*} \beta(E)=&\sup\{\lambda_d(C): C \subseteq E \text{ and } C \text{ is closed}\},\\ \gamma(E)=&\sup\{\lambda_d(K): K \subseteq E \text{ and } K \text{ is compact}\}. \end{align*}\]
Every compact subset of \(\mathbb R^d\) is closed, so \(\gamma(E) \leq \beta(E)\).
Fix a closed set \(C \subseteq E\) and put \(K_n=C \cap \overline{B(0, n)}\) for \(n \in \mathbb N\). Each \(K_n\) is closed and bounded, hence compact, and \(K_n \subseteq K_{n+1} \subseteq C\subseteq E\).
By continuity from below \(\lim_{n\to \infty}\lambda_d(K_n) = \lambda_d(C)\), so for every \(n \in \mathbb N\) we have \(\lambda_d(K_n)\le \gamma(E)\) and consequently \[\lambda_d(C) = \lim_{n\to \infty}\lambda_d(K_n) \leq \gamma(E).\]
Taking the supremum over all closed \(C \subseteq E\) we obtain \(\beta(E) \leq \gamma(E)\).
Hence \(\beta(E) = \gamma(E)\) and the proof is finished.
Theorem. If \(E \in \mathcal L(\mathbb R^d)\) and \(\lambda_d(E)<\infty\), then for every \(\varepsilon>0\) there is a set \(A\in \mathcal L(\mathbb R^d)\) which is a finite union of open rectangles such that \[\lambda_d(E \triangle A)<\varepsilon,\] where \(E \triangle A=(E \setminus A) \cup (A \setminus E)\) is the symmetric difference of \(E\) and \(A\).
In other words, every Lebesgue measurable set of finite measure is, up to a set of arbitrarily small measure, a finite union of open rectangles.
Proof: Let \(\varepsilon>0\). Since \(E \in \mathcal L(\mathbb R^d)\), by the previous theorem there is an open set \(U \supseteq E\) with \(\lambda_d(U \setminus E)<\frac{\varepsilon}{2}\). In particular \[\lambda_d(U) \leq \lambda_d(E)+\frac{\varepsilon}{2}<\infty.\]
Since \(U\) is open, \(U \in \mathcal L(\mathbb R^d)\), so by the previous theorem there is a compact set \(K \subseteq U\) with \(\lambda_d(U \setminus K)<\frac{\varepsilon}{2}\).
Every \(x \in K\) has an open rectangle \(R_x\) with \(x \in R_x \subseteq U\), and by compactness finitely many of them cover \(K\). Let \[A=R_{x_1} \cup \ldots \cup R_{x_m},\] so that \(K \subseteq A \subseteq U\) and, by monotonicity, \[\lambda_d(U \setminus A) \leq \lambda_d(U \setminus K)<\frac{\varepsilon}{2}.\]
Since \(E \subseteq U\) and \(A \subseteq U\) we have \(E \setminus A \subseteq U \setminus A\) and \(A \setminus E \subseteq U \setminus E\), hence \[\lambda_d(E \triangle A)=\lambda_d(E \setminus A)+\lambda_d(A \setminus E) \leq \lambda_d(U \setminus A)+\lambda_d(U \setminus E)<\varepsilon. \]
Theorem.
For every \(E \subseteq \mathbb R^d\) and \(x \in \mathbb R^d\) we have \[\lambda_d^{*}(x+E)=\lambda_d^{*}(E), \quad \text{ where } \quad x+E=\{x+y: y \in E\}.\]
For every \(E \in \mathcal L(\mathbb R^d)\) and \(x \in \mathbb R^d\) we have \(x+E \in \mathcal L(\mathbb R^d)\) and \[\lambda_d(x+E)=\lambda_d(E).\]
The key observation is that \(x+I \in J_{co}^d\) and \[{\rm v}_d(x+I)={\rm v}_d(I)\] for every \(I \in J_{co}^d\), since a translation shifts all the vertices of a rectangle by the same vector.
Proof of (i).
For every \(I \in J_{co}^d\) and \(x \in \mathbb R^d\) we have \(x+I \in J_{co}^d\) and \({\rm v}_d(x+I)={\rm v}_d(I)\).
Take \((I_n)_{n \in \mathbb N} \subseteq J_{co}^d\) so that \(E \subseteq \bigcup_{n \in \mathbb N}I_n\), then for any \(x \in \mathbb R^d\): \[x+E \subseteq x+\bigcup_{n \in \mathbb N}I_n=\bigcup_{n \in \mathbb N}(x+I_n).\]
Thus \[\lambda_d^{*}(x+E) \leq \sum_{n \in \mathbb N}{\rm v}_d(x+I_n)=\sum_{n \in \mathbb N}{\rm v}_d(I_n),\] and taking the infimum over all such coverings we obtain \(\lambda_d^{*}(x+E) \leq \lambda_d^{*}(E)\) for any \(E \subseteq \mathbb R^d\) and \(x \in \mathbb R^d\).
Applying this result with \(-x\) in place of \(x\) and \(x+E\) in place of \(E\) we obtain \[\lambda_d^{*}(E)=\lambda_d^{*}(-x+(x+E)) \leq \lambda_d^{*}(x+E).\]
Consequently we obtain \(\lambda_d^{*}(E)=\lambda_d^{*}(x+E)\) as desired.
Proof of (ii).
If \(E \in \mathcal L(\mathbb R^d)\) and \(x \in \mathbb R^d\) we show that \(x+E \in \mathcal L(\mathbb R^d)\).
Let \(A \subseteq \mathbb R^d\) and observe that, by part (i), we have \[\begin{align*} &\lambda_d^{*}(A \cap (x+E))+\lambda_d^{*}(A \cap (x+E)^c)\\ &=\lambda_d^{*}\big((A \cap (x+E))-x\big)+\lambda_d^{*}\big((A \cap (x+E)^c)-x\big)\\ &=\lambda_d^{*}((A-x) \cap E)+\lambda_d^{*}((A-x) \cap E^c)\\ &=\lambda_d^{*}(A-x)\\ &=\lambda_d^{*}(A), \end{align*}\] where we used that \((x+E)^c=x+E^c\) and that \(E \in \mathcal L(\mathbb R^d)\).
Thus \(x+E \in \mathcal L(\mathbb R^d)\) and \[\lambda_d(x+E)=\lambda_d(E),\] which completes the proof.
Theorem.
For every \(E \subseteq \mathbb R^d\) and \(\alpha \in \mathbb R\) we have \[\lambda_d^{*}(\alpha E)=|\alpha|^{d}\lambda_d^{*}(E), \quad \text{ where } \quad \alpha E=\{\alpha y: y \in E\}.\]
If \(E \in \mathcal L(\mathbb R^d)\) and \(\alpha \in \mathbb R\), then \(\alpha E \in \mathcal L(\mathbb R^d)\) and \[\lambda_d(\alpha E)=|\alpha|^{d}\lambda_d(E).\]
Recall from Lecture 4 that \(\lambda_d^{*}=\lambda_{d, o}^{*}\), so we may compute \(\lambda_d^{*}\) using coverings by open rectangles.
This is convenient here, since for \(\alpha \neq 0\) we have \(\alpha I \in J_{o}^d\) and \({\rm v}_d(\alpha I)=|\alpha|^{d}{\rm v}_d(I)\) for every \(I \in J_{o}^d\), whereas the family \(J_{co}^d\) is not preserved by \(\alpha<0\), because \(\alpha[a, b)=(\alpha b, \alpha a]\).
Proof of (i).
If \(\alpha=0\), then \(\alpha E \subseteq \{0\}\), thus \(\lambda_d^{*}(\alpha E)=0=|\alpha|^{d}\lambda_d^{*}(E)\), with the convention \(0 \cdot \infty=0\).
Let \(\alpha \neq 0\) and \(E \subseteq \mathbb R^d\). Let \[S=\{(I_n)_{n \in \mathbb N}: (I_n)_{n \in \mathbb N} \subseteq J_o^d\}.\]
Define a mapping \(M_{\alpha}:S \to S\) by \(M_{\alpha}(s)=(\alpha I_n)_{n \in \mathbb N}\) if \(s=(I_n)_{n \in \mathbb N}\).
\(M_{\alpha}\) is \(1-1\) and onto and its inverse is given by \(M_{1/\alpha}\).
Let \[S_E=\big\{(I_n)_{n \in \mathbb N} \in S: E \subseteq \bigcup_{n \in \mathbb N}I_n\big\}.\] Then \(M_{\alpha}[S_E]=S_{\alpha E}\), since \[E \subseteq \bigcup_{n \in \mathbb N}I_n \quad \iff \quad \alpha E \subseteq \bigcup_{n \in \mathbb N}\alpha I_n.\]
Define a set function \(\beta:S \to [0, \infty]\) by \(\beta(s)=\sum_{n \in \mathbb N}{\rm v}_d(I_n)\) for \(s=(I_n)_{n \in \mathbb N}\). Then \[\lambda_d^{*}(E)=\inf_{s \in S_E}\beta(s) \quad \text{ and } \quad \lambda_d^{*}(\alpha E)=\inf_{t \in S_{\alpha E}}\beta(t).\]
Moreover \[\beta(M_{\alpha}(s))=\sum_{n \in \mathbb N}{\rm v}_d(\alpha I_n) =|\alpha|^{d}\sum_{n \in \mathbb N}{\rm v}_d(I_n)=|\alpha|^{d}\beta(s).\]
Thus we obtain \(\lambda_d^{*}(\alpha E)=|\alpha|^{d}\lambda_d^{*}(E)\), since \(M_{\alpha}[S_E]=S_{\alpha E}\) and \[\lambda_d^{*}(\alpha E)=\inf_{t \in S_{\alpha E}}\beta(t)=\inf_{s \in S_E}\beta(M_{\alpha}(s)) =\inf_{s \in S_E}|\alpha|^{d}\beta(s)=|\alpha|^{d}\lambda_d^{*}(E)\] as desired.
Proof of (ii).
If \(\alpha=0\), then \(\alpha E \subseteq \{0\}\) is a null set, hence \(\alpha E \in \mathcal L(\mathbb R^d)\) by the completeness of \(\lambda_d\) and \(\lambda_d(\alpha E)=0=|\alpha|^{d}\lambda_d(E)\).
Let \(E \in \mathcal L(\mathbb R^d)\), \(\alpha \neq 0\) and \(A \subseteq \mathbb R^d\). We show that \(\alpha E \in \mathcal L(\mathbb R^d)\). Note that \[\lambda_d^{*}(\alpha^{-1}A)=\lambda_d^{*}(\alpha^{-1}A \cap E)+\lambda_d^{*}(\alpha^{-1}A \cap E^c).\]
Since \(\alpha^{-1}A \cap E=\alpha^{-1}(A \cap \alpha E)\) and \(\alpha^{-1}A \cap E^c=\alpha^{-1}(A \cap (\alpha E)^c)\), part (i) gives \[\frac{1}{|\alpha|^{d}}\lambda_d^{*}(A)=\frac{1}{|\alpha|^{d}}\lambda_d^{*}(A \cap \alpha E) +\frac{1}{|\alpha|^{d}}\lambda_d^{*}(A \cap (\alpha E)^c).\]
Hence \(\alpha E \in \mathcal L(\mathbb R^d)\), since \[\lambda_d^{*}(A)=\lambda_d^{*}(A \cap \alpha E)+\lambda_d^{*}(A \cap (\alpha E)^c).\]
By the previous part we also obtain \(\lambda_d(\alpha E)=|\alpha|^{d}\lambda_d(E)\).
The Lebesgue measure space \((\mathbb{R}^d,\mathcal{L}(\mathbb{R}^d),\lambda_d)\) is the completion of the Borel measure space \((\mathbb{R}^d,{\rm Bor}(\mathbb{R}^d),\lambda_d)\).
Proof. Let \(\overline{{\rm Bor}(\mathbb{R}^d)}\) be the completion of \({\rm Bor}(\mathbb{R}^d)\) with respect to \(\lambda_d\). i.e. \[\overline{{\rm Bor}(\mathbb{R}^d)}=\{A\cup B \subseteq \mathbb R^d: A\in {\rm Bor}(\mathbb{R}^d) \text{ and } B\subseteq C \in {\rm Bor}_0(\mathbb{R}^d)\},\] where \({\rm Bor}_0(\mathbb{R}^d)=\{C\in {\rm Bor}(\mathbb{R}^d): \lambda_d(C)=0 \}\). Since \((\mathbb{R}^d,\mathcal{L}(\mathbb{R}^d),\lambda_d)\) is complete and \({\rm Bor}(\mathbb{R}^d) \subseteq \mathcal{L}(\mathbb{R}^d)\) thus \[\overline{{\rm Bor}(\mathbb{R}^d)}\subseteq \mathcal{L}(\mathbb{R}^d).\]
Conversely, if \(E \in \mathcal{L}(\mathbb{R}^d)\) then \(E=F\cup N\), where \(F\) is an \(F_\sigma\) set and \(\lambda_d(N)=0\) thus \(E\in \overline{{\rm Bor}(\mathbb{R}^d)}\) and consequently \[\qquad \qquad \mathcal{L}(\mathbb{R}^d) \subseteq \overline{{\rm Bor}(\mathbb{R}^d)}. \qquad \]
Definition. Let \(F:\mathbb R\to \mathbb R\) be an increasing and right-continuous function. The Lebesgue–Stieltjes outer measure is defined for any \(E \subseteq \mathbb{R}\) by \[\begin{align*} \mu_{F}^{*}(E)=\inf\Big\{\sum_{n \in \mathbb{N}}(F(b_n)-F(a_n)): E \subseteq \bigcup_{n \in \mathbb{N}}(a_n, b_n]\Big\}. \end{align*}\]
The Lebesgue–Stieltjes measure \(\mu_F\) is the restriction of \(\mu_F^*\) to the Carathéodory \(\sigma\)-algebra \(\mathcal{L}_{\mu_F}=\mathcal{M}(\mu^{*}_F)\) of \(\mu^{*}_F\)-measurable sets.
\(\mathcal{L}_{\mu_F}=\mathcal{M}(\mu^{*}_F)\) will be called the Lebesgue–Stieltjes \(\sigma\)-algebra.
The members of \(\mathcal{L}_{\mu_F}\) are the Lebesgue–Stieltjes measurable sets.
By Carathéodory’s theorem \((\mathbb R, \mathcal{L}_{\mu_F}, \mu_F)\) is a complete measure space.
Moreover, \((\mathbb R, \mathcal{L}_{\mu_F}, \mu_F)\) is a \(\sigma\)-finite measure space and \({\rm Bor}(\mathbb{R})\subseteq\mathcal{L}_{\mu_F}\) and \(\mu_F((a, b])=F(b)-F(a)\) for any \(-\infty\le a\le b\le \infty\).
Problem 1. For \(E \subseteq \mathbb{R}\) prove that \(\mu_{F}^{*}(E)=\inf\big\{\sum_{n \in \mathbb{N}}\mu_F((a_n,b_n)): E \subseteq \bigcup_{n \in \mathbb{N}}(a_n, b_n)\big\}\).
Using this show that the following conditions are equivalent:
\(E \in \mathcal{L}_{\mu_F}\);
For every \(\varepsilon>0\) there exists an open set \(O \supseteq E\) so that \(\mu_F^{*}(O \setminus E) \leq \varepsilon\);
There exists a \(G_{\delta}\)-set \(G \supseteq E\) with \(\mu_F^{*}(G \setminus E)=0\);
For every \(\varepsilon>0\) there exists a closed set \(C \subseteq E\) with \(\mu_F^*(E \setminus C) \leq \varepsilon\);
There exists an \(F_{\sigma}\) set \(F \subseteq E\) with \(\mu_F^*(E \setminus F)=0\).
Moreover, if \(E \in \mathcal{L}_{\mu_F}\), then one has \[\begin{align*} \mu_F(E)&=\inf\{\mu_F(U): U\supseteq E \text{ and } U \text{ is open}\}\\ &=\sup\{\mu_F(F): F\subseteq E \text{ and } F \text{ is closed}\}\\ &=\sup\{\mu_F(K): K\subseteq E \text{ and } K \text{ is compact}\}. \end{align*}\]
Outer measures induced by set functions. Let \(X \neq\varnothing\) be a set and let \(\mathcal E \subseteq \mathcal P(X)\) be a collection containing \(\varnothing, X\). Let \(\rho:\mathcal E \to [0, \infty]\) be a set function such that \(\rho(\varnothing)=0\). For any \(A \subseteq X\)
\[\mu^{*}(A)=\inf\bigg\{\sum_{j \in \mathbb N}\rho(E_j): (E_j)_{j \in \mathbb N} \subseteq \mathcal E \text{ and } A \subseteq \bigcup_{j \in \mathbb N}E_j\bigg\}\] defines an outer measure, which we call an outer measure induced by \(\rho\).
By Carathéodory’s theorem the collection of \(\mu^{*}\)-measurable sets
\[\mathcal M(\mu^{*})=\big\{A \subseteq X: \mu^{*}(E)=\mu^{*}(E \cap A)+\mu^{*}(E \cap A^c) \ \text{ for all } \ E \subseteq X\big\}\] is a \(\sigma\)-algebra, which will be called the Carathéodory \(\sigma\)-algebra.
Moreover, \((X, \mathcal M(\mu^{*}), \mu^{*})\) is a complete measure space, and the outer measure \(\mu^{*}\) restricted to \(\mathcal M(\mu^{*})\) will be called the Carathéodory measure.
Definition. Let \(X\) be a set and let \(\mathcal E \subseteq \mathcal P(X)\). We define
\(\mathcal E_{\sigma}=\{\bigcup_{n \in \mathbb N}E_n: (E_n)_{n \in \mathbb N} \subseteq \mathcal E\}\) - the collection of countable unions of sets from \(\mathcal E\).
\(\mathcal E_{\delta}=\{\bigcap_{n \in \mathbb N}E_n: (E_n)_{n \in \mathbb N} \subseteq \mathcal E\}\) - the collection of countable intersections of sets from \(\mathcal E\).
\(\mathcal E_{\sigma\delta}=(\mathcal E_{\sigma})_{\delta}\) - the collection of countable intersections of sets from \(\mathcal E_{\sigma}\).
\(\mathcal E_{\delta\sigma}=(\mathcal E_{\delta})_{\sigma}\) - the collection of countable unions of sets from \(\mathcal E_{\delta}\).
\(\mathcal E^c=\{E^c: E \in \mathcal E\}\) - the collection of complements of sets from \(\mathcal E\).
Note that \((\mathcal E_{\sigma})^c=(\mathcal E^c)_{\delta}\) and \((\mathcal E_{\delta})^c=(\mathcal E^c)_{\sigma}\).
Problem 2. Let \(X\), \(\mathcal E\), \(\rho\) and \(\mu^{*}\) be as above. Prove that \(\mu^{*}\) is an outer measure on \(X\). Where do we use the assumption that \(\varnothing, X \in \mathcal E\)?
Problem 3. Let \(\mathcal E \subseteq \mathcal P(X)\). Prove that \[(\mathcal E_{\sigma})^c=(\mathcal E^c)_{\delta}, \quad (\mathcal E_{\delta})^c=(\mathcal E^c)_{\sigma} \quad \text{ and } \quad (\mathcal E_{\sigma\delta})^c=(\mathcal E^c)_{\delta\sigma}.\]
The Lebesgue outer measure \(\lambda_d^{*}\) is the outer measure induced by the volume function \({\rm v}_d\) on \(\mathcal E=J_{o}^d\); in this case
\(\mathcal E_{\sigma\delta}\) plays the role of the \(G_{\delta}\)-sets;
\((\mathcal E^c)_{\delta\sigma}\) the role of the \(F_{\sigma}\)-sets.
Assumptions. Let \(X \neq\varnothing\) be a set and let \(\mathcal E \subseteq \mathcal P(X)\) be a collection containing \(\varnothing, X\). Let \(\rho:\mathcal E \to [0, \infty]\) be a set function such that \(\rho(\varnothing)=0\), and for \(A \subseteq X\) let
\[\mu^{*}(A)=\inf\bigg\{\sum_{j \in \mathbb N}\rho(E_j): (E_j)_{j \in \mathbb N} \subseteq \mathcal E \text{ and } A \subseteq \bigcup_{j \in \mathbb N}E_j\bigg\}.\] Let \(\mathcal M(\mu^{*})\) be the Carathéodory \(\sigma\)-algebra. In the problems below we assume, as indicated in each of them, that
(a) \(\mu^{*}=\rho\) on \(\mathcal E\);
(b) \(\mathcal E \subseteq \mathcal M(\mu^{*})\);
(c) \(\mu^{*}\) is a \(\sigma\)-finite measure on \((X, \mathcal M(\mu^{*}))\).
Problem 4. Suppose that (a) holds. Prove that
(i) for any \(A \subseteq X\) and \(\varepsilon>0\) there exists a set \(U \supseteq A\) such that
\[U \in \mathcal E_{\sigma} \quad \text{ and } \quad \mu^{*}(A) \leq \mu^{*}(U) \leq \mu^{*}(A)+\varepsilon;\]
(ii) for any \(A \subseteq X\) there exists a set \(B \supseteq A\) such that
\[B \in \mathcal E_{\sigma\delta} \quad \text{ and } \quad \mu^{*}(A)=\mu^{*}(B).\]
Problem 5. Suppose that (a) holds. Prove that \(\mu^{*}\) is outer regular with respect to the family \(\mathcal E_{\sigma}\), that is, for every \(E \subseteq X\) one has
\[\mu^{*}(E)=\inf\{\mu^{*}(U): U \supseteq E \text{ and } U \in \mathcal E_{\sigma}\}.\]
Problem 6. Suppose that (a), (b) and (c) hold. Prove that for \(E \subseteq X\) the following conditions are equivalent:
(i) \(E \in \mathcal M(\mu^{*})\);
(ii) for every \(\varepsilon>0\) there exists a set \(U \supseteq E\) such that
\[U \in \mathcal E_{\sigma} \quad \text{ and } \quad \mu^{*}(U \setminus E) \leq \varepsilon;\]
(iii) there exists a set \(G \supseteq E\) such that
\[G \in \mathcal E_{\sigma\delta} \quad \text{ and } \quad \mu^{*}(G \setminus E)=0;\]
(iv) for every \(\varepsilon>0\) there exists a set \(C \subseteq E\) such that
\[C \in (\mathcal E^c)_{\delta} \quad \text{ and } \quad \mu^{*}(E \setminus C) \leq \varepsilon;\]
(v) there exists a set \(F \subseteq E\) such that
\[F \in (\mathcal E^c)_{\delta\sigma} \quad \text{ and } \quad \mu^{*}(E \setminus F)=0.\]
Problem 7. Suppose that assumptions (a), (b) and (c) hold. Prove that
(i) \(\mu^{*}\) is outer regular with respect to the family \(\mathcal E_{\sigma}\), that is,
\[\mu^{*}(E)=\inf\{\mu^{*}(U): U \supseteq E \text{ and } U \in \mathcal E_{\sigma}\} \quad \text{ for any } \quad E \in \mathcal M(\mu^{*});\]
(ii) \(\mu^{*}\) is inner regular with respect to the family \((\mathcal E^c)_{\delta}\), that is,
\[\mu^{*}(E)=\sup\{\mu^{*}(C): C \subseteq E \text{ and } C \in (\mathcal E^c)_{\delta}\} \quad \text{ for any } \quad E \in \mathcal M(\mu^{*}).\]
Compare with the regularity theorem for \(\lambda_d\) proved earlier in this lecture.
Problem 8. Suppose that assumptions (a) and (b) hold and that \((X, \mathcal M(\mu^{*}), \mu^{*})\) is a finite measure space. Prove that for every \(E \in \mathcal M(\mu^{*})\) and every \(\varepsilon>0\) there exists a set \(A \in \mathcal E_{\sigma}\) such that \[\mu^{*}(E \triangle A)<\varepsilon.\] In fact one can take \(A\) to be a finite union of elements of \(\mathcal E\), which is the important part of the statement.
For \(X=\mathbb R^d\), \(\mathcal E=J_{o}^d\) and \(\rho={\rm v}_d\) this is the theorem on approximation of sets of finite measure by finite unions of rectangles.
We now deduce an analogue for the Lebesgue–Stieltjes measures:
If \(E\in \mathcal L_{\mu_F}\) and \(\mu_F(E)<\infty\), then for every \(\varepsilon>0\) there is a set \(A\) that is a finite union of open intervals such that \(\mu_F(E\triangle A)<\varepsilon.\)
Problem 9. Let \((X, \mathcal M, \mu)\) be a measure space and let \(\mathcal E \subseteq \mathcal M\) be a collection containing \(\varnothing, X\). Consider the following two conditions:
(a) for every \(E \in \mathcal M\) and every \(\varepsilon>0\) there exists a set \(G \in \mathcal E\) such that
\[G \supseteq E \quad \text{ and } \quad \mu(G \setminus E) \leq \varepsilon;\]
(b) \(\mu\) is outer regular with respect to the family \(\mathcal E\), i.e.
\[\mu(E)=\inf\{\mu(U): U \supseteq E \text{ and } U \in \mathcal E\} \ \text{ for any } \ E \in \mathcal M.\]
Prove that
(i) (a) implies (b);
(ii) if \(\mu(X)<\infty\), then (b) implies (a);
(iii) if \(\mu\) is \(\sigma\)-finite and \(\mathcal E\) is closed under countable unions, then (b) implies (a).
Problem 10. Let \((X, \mathcal M, \mu)\) be a measure space and let \(\mathcal E \subseteq \mathcal M\) be a collection containing \(\varnothing, X\). Consider the following two conditions:
(a) for every \(E \in \mathcal M\) and every \(\varepsilon>0\) there exists a set \(F \in \mathcal E\) such that
\[F \subseteq E \quad \text{ and } \quad \mu(E \setminus F) \leq \varepsilon;\]
(b) \(\mu\) is inner regular with respect to the family \(\mathcal E\), i.e.
\[\mu(E)=\sup\{\mu(C): C \subseteq E \text{ and } C \in \mathcal E\} \ \text{ for any } \ E \in \mathcal M.\]
Prove that
(i) (a) implies (b);
(ii) if \(\mu(X)<\infty\), then (b) implies (a);
(iii) if \(\mu\) is \(\sigma\)-finite and \(\mathcal E\) is closed under countable unions, then (b) implies (a).
Problem 11. Let \((X, \mathcal M, \mu)\) be a \(\sigma\)-finite measure space and let \(\mathcal E \subseteq \mathcal M\) be a collection containing \(\varnothing, X\). Prove that
(i) if \(\mu\) is outer regular with respect to \(\mathcal E\) and \(\mathcal E\) is closed under countable unions, then for every \(E \in \mathcal M\) there exists a set \(G \in \mathcal E_{\delta}\) such that
\[G \supseteq E \quad \text{ and } \quad \mu(G \setminus E)=0;\]
(ii) if \(\mu\) is inner regular with respect to \(\mathcal E\), then for every \(E \in \mathcal M\) there exists a set \(F \in \mathcal E_{\sigma}\) such that
\[F \subseteq E \quad \text{ and } \quad \mu(E \setminus F)=0.\]
Note the asymmetry: in (ii) no closure assumption on \(\mathcal E\) is needed.