Let \(J_o\) be the collection of all open intervals in \(\mathbb{R}\): \[J_o=\{(a,b): -\infty \leq a<b \leq \infty\} \cup \{\varnothing\}.\]
Let \(J_{co}\) be the collection of all closed-open intervals in \(\mathbb{R}\): \[J_{co}=\{[a,b): -\infty \leq a<b \leq \infty\} \cup \{\varnothing\}.\]
Let \(J_{oc}\) be the collection of all open-closed intervals in \(\mathbb{R}\): \[J_{oc}=\{(a,b]: -\infty \leq a<b \leq \infty\} \cup \{\varnothing\}.\]
Let \(J_{c}\) be the collection of all closed intervals in \(\mathbb{R}\): \[J_c=\{[a,b]: -\infty \leq a<b \leq \infty\} \cup \{\varnothing\}.\]
Let \(J=J_o \cup J_{oc} \cup J_{co} \cup J_c\) be the collection of all intervals in \(\mathbb{R}\).
\(F\) is increasing, since \(\mu\) is monotone.
\(F\) is right continuous, since \(\bigcap_{n\in\mathbb N} (-\infty, x_n]=(-\infty, x]\) and \[\lim_{n\to \infty}F(x_n)=\lim_{n\to \infty}\mu((-\infty, x_n])=\mu((-\infty, x])=F(x)\] whenever \(x_n\searrow x\) as \(n\to \infty\).
Moreover, if \(-\infty<a<b\le\infty\) then \((-\infty, b]=(-\infty, a]\cup (a, b]\) thus \[\mu((a, b])=F(b)-F(a),\] where \((a, \infty]=(a, \infty)\) and \(F(\infty)=\lim_{x\to \infty}F(x)\).
Our procedure will be to turn this process around and construct a measure \(\mu\) starting from an increasing, right-continuous function \(F\). As a particular case, if \(F(x)=x\) we construct the Lebesgue measure on \(\mathbb R\).
Theorem. (From countably subadditive set functions to premeasures) Let \(\rho:\mathcal{E} \to [0,\infty]\) be a set function on a semi-algebra \(\mathcal{E}\) of subsets of a set \(X\) and \(\rho(\varnothing)=0\). Suppose that \(\rho\) is finitely additive on \(\mathcal{E}\) and let \(\mathcal{A}\) be the algebra that consists of all finite disjoint unions of members of \(\mathcal{E}\). Define a set function \(\mu_0\) on \(\mathcal{A}\) by setting \[\mu_0(A)=\sum_{i=1}^{n}\rho(E_i)\] for \(A=\bigcup_{j=1}^{n}E_j \in \mathcal{A}\), where \(E_1,\ldots,E_n \in \mathcal{E}\) and \(E_i \cap E_j = \varnothing\) if \(i \neq j\).
Then \(\mu_0\) is a well-defined additive set function on \(\mathcal{A}\) such that \(\mu_0(\varnothing)=0\) and \(\mu_0=\rho\) on \(\mathcal{E}\).
If additionally \(\rho\) is countably subadditive on \(\mathcal{E}\) then \(\mu_0\) is a premeasure on \(\mathcal{A}\).
A simple fact \(J_{oc}\) and \(J_{co}\) are semi-algebras.
Indeed, note that
an intersection of two half-open intervals is again a half-open interval.
the complement of a half-open interval is a half-open interval or the disjoint union of two half-open intervals.
Set functions on \(J_{oc}\) induced by increasing and right-continuous functions Let \(F:\mathbb R\to \mathbb R\) be an increasing and right-continuous function. We define a set function \(\rho_F:J_{oc} \to [0,\infty]\) by setting \(\rho_F(\varnothing)=0\) and \[\rho_F(I)= F(b)-F(a) \quad\text{ whenever } \quad I=(a, b]\in J_{oc},\] with the understanding that \((a, \infty]=(a, \infty)\) and \(F(\infty)=\lim_{x\to \infty}F(x)\).
Remark.
Let \(\mathcal{A}\) be the algebra of finite disjoint unions of elements from \(J_{oc}\). Then it is clear that \[\sigma(\mathcal{A})=\sigma(J_{oc})={\rm { Bor}}(\mathbb{R})=\mathcal B_{\mathbb R}.\]
It is easy to see that \(\rho_F\) is finitely additive on the semi-algebra \(J_{oc}\), i.e. for any disjoint family \((I_j)_{j=1}^n \subseteq J_{oc}\) such that \(\bigcup_{j=1}^nI_j\in J_{oc}\) we have \[\rho_F\bigg(\bigcup_{j=1}^{n}I_j\bigg)=\sum_{j=1}^{n}\rho_F(I_j).\]
A finitely additive set function on \(\mathcal A\). Define a set function \(\mu_{F, 0}\) on \(\mathcal{A}\) by setting \[\mu_{F, 0}(E)=\sum_{i=1}^{n}\rho_F(I_i)\] for \(E=\bigcup_{j=1}^{n}I_j \in \mathcal{A}\), where \((I_j)_{j=1}^n\subseteq J_{oc}\) and \(I_i \cap I_j = \varnothing\) if \(i \neq j\).
By the previous theorem and last remark \(\mu_{F, 0}\) is well defined and finitely additive on \(\mathcal{A}\).
Our goal is to show that \(\rho_F\) is countably subadditive on \(J_{oc}\). Then by the previous theorem we conclude that \(\rho_F\) is countably additive on \(J_{oc}\), which in turn implies that its extension \(\mu_{F, 0}\) to \(\mathcal{A}\) is also countably additive on \(\mathcal{A}\). In other words, \(\mu_{F, 0}\) is a premeasure on \(\mathcal{A}\).
Proof. We have to prove that for any \((I_n)_{n \in \mathbb{N}} \subseteq J_{oc}\) such that \(I=\bigcup_{n \in \mathbb{N}}I_n\in J_{oc}\) we have \[\rho_F(I) \leq \sum_{n \in \mathbb{N}}\rho_F(I_n).\] We first assume that \(\rho_F(I)<\infty\).
If \(I =\varnothing\) then of course \(\rho_F(I)=0 \leq \sum_{n \in \mathbb{N}}\rho_F(I_n)\).
If \(\rho_F(I_n)=\infty\) for some \(n \in \mathbb{N}\), then there is nothing to prove, since \[\rho_F(I)<\infty=\sum_{n \in \mathbb{N}}\rho_F(I_n).\]
We may assume that \(\rho_F(I_n)<\infty\) for all \(n \in \mathbb{N}\).
Let \(I=(a,b]\) for some \(a<b\) and let \(H_x=(x, \infty)\) and define \[A=\Big\{x \in [a, b]: F(b)-F(x) \leq \sum_{j \in \mathbb{N}}\rho_F(I_j \cap H_{x})\Big\}.\]
Note that if \(I_j=(c_j,d_j]\) then \(I_j \cap H_x=(\max\{c_j, x\},d_j]\), so the quantity \(\rho_F(I_j \cap H_x)=F(d_j)-F(\max\{c_j, x\})\) is always defined.
We have that \(b \in A\) since \[F(b)-F(b)=0 \leq \sum_{j \in \mathbb{N}}\rho_F(I_j \cap H_x) \quad \text{ and } \quad A \subseteq [a,b].\]
Thus \(c=\inf A\) is defined and \(c=\inf A \in [a,b]\). We show that \(c \in A\). Indeed, right-continuity of \(F\) implies \[F(b)-F(c)=\inf_{x \in A}(F(b)-F(x)) \leq \inf_{x \in A}\sum_{j \in \mathbb{N}}\rho_F(I_j \cap H_x) \leq \sum_{j \in \mathbb{N}}\rho_F(I_j \cap H_c),\] since \(\rho_F(I_j \cap H_x)\le \rho_F(I_j \cap H_c)\) as \(c\le x\in A\).
We now prove that \(c=a\). If \(a<c\), then \(c \in (a,b]\) so there is \(k \in \mathbb{N}\) such that \(c \in I_k=(c_k,d_k]\), then \(x=\max\{c_k,a\}<c\le b\).
For each \(j \in \mathbb{N}\) we have \[\rho_F(I_j \cap H_c) \leq \rho_F(I_j \cap H_x)\] while \[\rho_F(I_k \cap H_x)=\rho_F(I_k \cap H_c)+(F(c)-F(x))\] and consequently \(x\in A\), since \[\begin{aligned} \sum_{j \in \mathbb{N}}\rho_F(I_j \cap H_x) &\geq \sum_{j \in \mathbb{N}}\rho_F(I_j \cap H_c)+(F(c)-F(x)) \\ &\geq (F(b)-F(c))+(F(c)-F(x))=F(b)-F(x). \end{aligned}\]
But \(x<c=\inf A\), which is impossible. Thus \(a=c \in A\), so \[\rho_F(I)=F(b)-F(a) \leq \sum_{j \in \mathbb{N}}\rho_F(I_j \cap H_c) \leq \sum_{j \in \mathbb{N}}\rho_F(I_j).\]
Suppose now that \(\ell(I)=\infty\).
Assume that \(I=(a,\infty)=\bigcup_{n \in \mathbb{N}}I_n\in J_{oc}\) for some \(a \in \mathbb{R}\). For any \(n \in \mathbb{N}\) such that \(n>a\) we have \((a,n] \subseteq (a,\infty)\) and \[\rho_F((a,n])\le \rho_F((a,\infty))\] then \[F(n)-F(a) \leq \sum_{n \in \mathbb{N}}\rho_F(I_n).\]
Passing with \(n \to \infty\) we get \[\rho_F(I)=\infty \leq \sum_{n \in \mathbb{N}}\rho_F(I_n).\]
This completes the proof of the theorem. $$\tag*{$\blacksquare$}$$
The Lebesgue–Stieltjes measure \(\mu_F\) is the restriction of \(\mu_F^*\) to the Carathéodory \(\sigma\)-algebra \(\mathcal{M}_{\mu_F}=\mathcal{M}(\mu^{*}_F)\) of \(\mu^{*}_F\)-measurable sets.
\(\mathcal{M}_{\mu_F}=\mathcal{M}(\mu^{*}_F)\) will be called the Lebesgue–Stieltjes \(\sigma\)-algebra.
The members of \(\mathcal{M}_{\mu_F}\) are the Lebesgue–Stieltjes measurable sets.
By Carathéodory’s theorem \((\mathbb R, \mathcal{M}_{\mu_F}, \mu_F)\) is a complete measure space.
The outer measure \(\mu_F^{*}\) is induced by the premeasure \(\mu_{F, 0}\) on the algebra \(\mathcal A\) of finite disjoint unions of elements from \(J_{oc}\). The premeasure \(\mu_{F, 0}\) is \(\sigma\)-finite, since \(\mathbb R=\bigcup_{n\in\mathbb N}(n, n+1]\).
\((\mathbb R, \mathcal{M}_{\mu_F}, \mu_F)\) is a complete \(\sigma\)-finite measure space.
By the Carathéodory extension theorem we conclude that
\(\mu_F(E)=\mu_{F}^{*}(E)=\mu_{F, 0}(E)\) for any \(E \in \mathcal{A}\);
\(\mathcal{A} \subseteq \sigma(J_{oc})=\sigma(\mathcal{A})=\mathcal{B}_{\mathbb{R}} \subseteq \mathcal{M}_{\mu_F}\).
Therefore it follows that \(\mu_F\) is a \(\sigma\)-finite Borel measure and \[\mu((a, b])=F(b)-F(a)\] for any \(-\infty\le a\le b\le \infty\).
Moreover \(\mu_F\) is unique in the sense that if \(G\) is another such function, we have \(\mu_F=\mu_G\) if and only if \(F-G\) is constant.
Conversely, if \(\mu\) is a Borel measure on \(\mathbb R\) that is finite on all bounded Borel sets and we define \[F(x)=\ \begin{cases} \mu((-\infty, x]) & \text{ if } x>0,\\ 0 & \text{ if } x=0,\\ -\mu((x, 0])& \text{ if } x<0,\\ \end{cases}\] then \(F\) is increasing and right continuous and \(\mu=\mu_F\).
For any \(E\in \mathcal{M}_{\mu_F}\) one has \[\mu_{F}(E)=\inf\Big\{\sum_{n \in \mathbb{N}}\mu_F((a_n, b_n)): E \subseteq \bigcup_{n \in \mathbb{N}}(a_n, b_n)\Big\}.\]
Proof. For any \(E\in \mathcal{M}_{\mu_F}\) we have to show that \(\mu_F(E)=\alpha_E\), where \[\alpha_E=\inf\Big\{\sum_{n \in \mathbb{N}}\mu_F((a_n, b_n)): E \subseteq \bigcup_{n \in \mathbb{N}}(a_n, b_n)\Big\}.\]
Suppose \(E\subseteq \bigcup_{j\in\mathbb N}(a_j, b_j)\), then \[\mu_F(E)\le \sum_{j\in \mathbb N}\mu_F((a_j, b_j)).\] Thus \(\mu_F(E)\le \alpha_E\).
Conversely, we have to show that \(\alpha_E\le \mu_F(E).\)
Given \(\varepsilon>0\) there exists a sequence \(((a_j, b_j])_{j\in\mathbb N}\subseteq J_{oc}\) with \[E\subseteq \bigcup_{j\in\mathbb N}(a_j, b_j] \quad \text{ and } \quad \sum_{j\in\mathbb N}\mu_F((a_j, b_j])\le \mu_F(E)+\varepsilon/2.\]
For each \(j\in \mathbb N\) there exists \(\delta_j>0\) such that \[\mu_F((b_j, b_j+\delta_j])=F(b_j+\delta_j)-F(b_j)<\varepsilon2^{-j-1}\] Then \(E\subseteq \bigcup_{j\in\mathbb N}(a_j, b_j+\delta_j)\) giving \(\alpha_E\le \mu_F(E)\), since \[\begin{gathered} \alpha_E\le \sum_{j\in\mathbb N}\mu_F((a_j, b_j+\delta_j)) \le \sum_{j\in\mathbb N}\mu_F((a_j, b_j])+\mu_F((b_j, b_j+\delta_j])\\ \le \sum_{j\in\mathbb N}\mu_F((a_j, b_j])+\varepsilon/2\le \mu_F(E)+\varepsilon. \blacksquare \end{gathered}\]
\(E \in \mathcal{M}_{\mu_F}\);
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 \(\mu_F^{*}(E \setminus F)=0\).
Proof. By the Carathéodory extension theorem we conclude that
\(\mu_F(E)=\mu_{F}^{*}(E)=\mu_{F, 0}(E)\) for any \(E \in \mathcal{A}\);
\(\mathcal{A} \subseteq \sigma(J_{oc})=\sigma(\mathcal{A})=\mathcal{B}_{\mathbb{R}} \subseteq \mathcal{M}_{\mu_F}\).
By the previous lemma for \(E\in\mathcal M_{\mu_F}\) we also have \[\begin{aligned} \mu_{F}(E)&= \inf\Big\{\sum_{n \in \mathbb{N}}(F(b_n)-F(a_n)): E \subseteq \bigcup_{n \in \mathbb{N}}(a_n, b_n]\Big\}\\ &=\inf\Big\{\sum_{n \in \mathbb{N}}\mu_F((a_n, b_n)): E \subseteq \bigcup_{n \in \mathbb{N}}(a_n, b_n)\Big\}. \end{aligned}\]
Using the regularity conditions for Carathéodory’s measures discussed in Lecture 4 we obtain the regularity results for the measure \(\mu_F\). $$\tag*{$\blacksquare$}$$
If \(E \in \mathcal{M}_{\mu_F}\) then \[\begin{aligned} \mu_F(E)&=\inf\{\mu_F(U): U\supseteq E \text{ and } U \text{ is open}\}\\ &=\sup\{\mu_F(F): F\subseteq E \text{ and } F \text{ is closed}\}\\ &=\sup\{\mu_F(K): K\subseteq E \text{ and } K \text{ is compact}\}. \end{aligned}\]
Proof. Since every open set in \(\mathbb R\) is \(\sigma\)-compact thus the measure \(\mu_F\) is a Radon regular measure on \(\mathbb R\), which is a consequence of the regularity results discussed in Lecture 5 and the previous theorem. $$\tag*{$\blacksquare$}$$
If \(E\in \mathcal M_{\mu_F}\) 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 \(\mu_F(E\triangle A)<\varepsilon.\)
Proof. By the Carathéodory extension theorem we conclude that
\(\mu_F(E)=\mu_{F}^{*}(E)=\mu_{F, 0}(E)\) for any \(E \in \mathcal{A}\);
\(\mathcal{A} \subseteq \sigma(J_{oc})=\sigma(\mathcal{A})=\mathcal{B}_{\mathbb{R}} \subseteq \mathcal{M}_{\mu_F}\).
By the previous lemma for \(E\in\mathcal M_{\mu_F}\) we also have \[\begin{aligned} \mu_{F}(E)&= \inf\Big\{\sum_{n \in \mathbb{N}}(F(b_n)-F(a_n)): E \subseteq \bigcup_{n \in \mathbb{N}}(a_n, b_n]\Big\}\\ &=\inf\Big\{\sum_{n \in \mathbb{N}}\mu_F((a_n, b_n)): E \subseteq \bigcup_{n \in \mathbb{N}}(a_n, b_n)\Big\}. \end{aligned}\]
Using the approximation theorem for Carathéodory’s measures discussed in Lecture 4 we obtain the desired approximation results for \(\mu_F\). $$\tag*{$\blacksquare$}$$
Definition. For \(I \in J\) we define the length function \(\ell:J \to [0,\infty]\) by \(\ell(\varnothing)=0\) and \[\ell(I)=\begin{cases} b-a &\text{ if } {\rm cl}(I)=[a, b] \text{ for some } a,b \in \mathbb{R},\\ \infty &\text{ otherwise}. \end{cases}\]
Remark. The length function \(\ell\) is a special case of the set function \(\rho_F\) on \(J_{oc}\) that corresponds to \(F(x)=x\) for \(x\in\mathbb R\). Therefore the length function \(\ell\) is countably additive on \(J_{oc}\) and defines a countably additive premeasure on the algebra \(\mathcal A\) of finite disjoint unions of elements from \(J_{oc}\).
The Lebesgue measure \(\lambda\) is the restriction of \(\lambda^*\) to the Carathéodory \(\sigma\)-algebra \(\mathcal{L}(\mathbb R)=\mathcal{M}(\lambda^{*})\) of \(\lambda^{*}\)-measurable sets.
\(\mathcal{L}(\mathbb R)=\mathcal{M}(\lambda^{*})\) will be called the Lebesgue \(\sigma\)-algebra.
The members of \(\mathcal{L}(\mathbb R)\) are the Lebesgue measurable sets.
By Carathéodory’s theorem \((\mathbb R, \mathcal{L}(\mathbb R), \lambda)\) is a complete measure space.
By Carathéodory’s extension theorem \(\mathcal B_{\mathbb R}\subseteq \mathcal{L}(\mathbb R)\) and \(\lambda(I)=\ell(I)\) for any \(I\in J=J_o \cup J_{oc} \cup J_{co} \cup J_c\) and \(\lambda\) is \(\sigma\)-finite on \(\mathbb R\).
\(E \in \mathcal L(\mathbb{R})\);
For every \(\varepsilon>0\) there exists an open set \(O \supseteq E\) so that \(\lambda^{*}(O \setminus E) \leq \varepsilon\);
There exists a \(G_{\delta}\)-set \(G \supseteq E\) with \(\lambda^{*}(G \setminus E)=0\);
For every \(\varepsilon>0\) there exists a closed set \(C \subseteq E\) with \(\lambda^*(E \setminus C) \leq \varepsilon\);
There exists an \(F_{\sigma}\) set \(F \subseteq E\) with \(\lambda^*(E \setminus F)=0\).
Moreover, if \(E \in \mathcal L(\mathbb{R})\), then one has \[\begin{aligned} \lambda(E)&=\inf\{\lambda(U): U\supseteq E \text{ and } U \text{ is open}\}\\ &=\sup\{\lambda(F): F\subseteq E \text{ and } F \text{ is closed}\}\\ &=\sup\{\lambda(K): K\subseteq E \text{ and } K \text{ is compact}\}. \end{aligned}\]
If \(E\in \mathcal L(\mathbb R)\) and \(\lambda(E)<\infty\), then for every \(\varepsilon>0\) there is a set \(A\) that is a finite union of open intervals such that \[\lambda(E\triangle A)<\varepsilon.\]
For every \(E \subseteq \mathbb{R}\) and \(x \in \mathbb{R}\) we have \[\lambda^{*}(x+E)=\lambda^{*}(E),\] where \(x+E=\{x+y:y \in E\}\).
For every \(E \in \mathcal{L}(\mathbb{R})\) and \(x \in \mathbb{R}\) we have \(x+E \in \mathcal{L}(\mathbb{R})\) and \[\lambda(x+E)=\lambda(E).\]
Proof of (i).
For every \(I \in J_{o}\) and \(x \in \mathbb{R}\) we have \(x+I \in J_{o}\) and \(\ell(x+I)=\ell(I)\).
Take \((I_n)_{n \in \mathbb{N}}\) so that \(E \subseteq \bigcup_{n \in \mathbb{N}}I_n\), then for any \(x \in \mathbb{R}\): \[x+E \subseteq x+\bigcup_{n \in \mathbb{N}}I_n=\bigcup_{n \in \mathbb{N}}(x+I_n).\]
Thus \[\lambda^{*}(x+E) \leq \sum_{n \in \mathbb{N}}\ell(x+I_n)=\sum_{n \in \mathbb{N}}\ell(I_n),\] hence \(\lambda^{*}(x+E)\le \lambda^{*}(E)\) for any \(E \subseteq \mathbb{R}\) and \(x \in \mathbb{R}\).
Applying this result with \(-x\) in place of \(x\) and \(x+E\) in place of \(E\) we obtain \[\lambda^{*}(E)=\lambda^*(-x+(x+E)) \leq \lambda^{*}(x+E),\]
Consequently we obtain \(\lambda^{*}(E)=\lambda^{*}(x+E)\) as desired.$$\tag*{$\blacksquare$}$$
Proof of (ii).
If \(E \in \mathcal{L}(\mathbb{R})\) and \(x \in \mathbb{R}\) we show that \(x +E \in \mathcal{L}(\mathbb{R})\).
Let \(A \subseteq \mathbb{R}\) and observe \[\begin{aligned} &\lambda^{*}(A \cap (x+E))+\lambda^{*}(A \cap (x+E)^c)\\ &=\lambda^{*}(A \cap (x+E)-x)+\lambda^{*}(A \cap (x+E)^c-x)\\ &=\lambda^{*}((A-x) \cap E)+\lambda^{*}((A-x) \cap E^c)\\ &=\lambda^{*}(A-x)\\ &=\lambda^{*}(A), \end{aligned}\] thus \(x+E \in \mathcal{L}(\mathbb{R})\) and \[\lambda(x+E)=\lambda(E),\] which completes the proof. $$\tag*{$\blacksquare$}$$
For \(E \subseteq \mathbb{R}\) and \(\alpha \in \mathbb{R}\) we have \[\lambda^{*}(\alpha E)=|\alpha|\lambda^{*}(E),\]
where \(\alpha E=\{\alpha x: x \in E\}\).
If \(E \in \mathcal{L}(\mathbb{R})\) and \(\alpha \in \mathbb{R}\) then \(\alpha R \in \mathcal{L}(\mathbb{R})\) and \[\lambda(\alpha E)=|\alpha|\lambda(E).\]
Proof of (i).
If \(\alpha=0\) then \(\alpha E=\{0\}\) thus \(\lambda^{*}(\alpha E)=0=|\alpha|\lambda^{*}(E)\).
Let \(\alpha \neq 0\) and \(E \subseteq \mathbb{R}\). Let \(S=\{(I_n)_{n\in\mathbb N}:(I_n)_{n\in\mathbb N}\subseteq J_o\}\).
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 \(\beta(s)=\sum_{n \in \mathbb{N}}\ell(I_n)\) for \(s=(I_{n})_{n \in \mathbb{N}}\). Then \[\lambda^{*}(E)=\inf_{s \in S_E}\beta(s) \quad \text{ and } \quad \lambda^{*}(\alpha E)=\inf_{t \in \alpha S_E}\beta(t).\] Then \[\beta(M_{\alpha}(s))=\sum_{n \in \mathbb{N}}\ell(\alpha I_n)=|\alpha|\sum_{n \in \mathbb{N}}\ell(I_n)=|\alpha|\beta(s).\] Thus we obtain \(\lambda^{*}(\alpha E)=|\alpha|\lambda^{*}(E)\), since \(M_{\alpha}[S_E]=S_{\alpha E}\) and \[\lambda^{*}(\alpha E)=\inf_{t \in S_{\alpha E}}\beta(t)=\inf_{s \in S_E}\beta(M_{\alpha}(s))=\inf_{s \in S_E}|\alpha|\beta(s)=|\alpha|\lambda^{*}(E). \tag*{$\blacksquare$}\]
Proof of (ii).
Let \(E \in \mathcal{L}(\mathbb{R})\) and \(\alpha \neq 0\). We show that \(\alpha E \in \mathcal{L}(\mathbb{R})\).
Note that \[\lambda^{*}({\alpha}^{-1}A)=\lambda^{*}({\alpha}^{-1}A \cap E)+\lambda^{*}({\alpha}^{-1}A \cap E^c).\]
Thus \[\frac{1}{|\alpha|}\lambda^{*}(A)=\frac{1}{|\alpha|}\lambda^{*}\left(A \cap \alpha E\right) +\frac{1}{|\alpha|}\lambda^{*}\left(A \cap (\alpha E)^c\right).\]
Hence \(\alpha E \in \mathcal{L}(\mathbb{R})\), since \[\lambda^{*}(A)=\lambda^{*}(A \cap \alpha E)+\lambda^{*}(A \cap (\alpha E)^c).\]
By the previous part we also obtain \(\lambda(\alpha E)=|\alpha|\lambda(E)\).$$\tag*{$\blacksquare$}$$