Definition of an outer measure. An outer measure on a set \(X\neq\varnothing\) is a function \(\mu^{*}:\mathcal{P}(X) \to [0,\infty]\) that satisfies
\(\mu^{*}(\varnothing)=0\),
\(\mu^{*}(A) \leq \mu^{*}(B)\) if \(A \subseteq B\),
\(\mu^{*}\left(\bigcup_{n \in \mathbb{N}}A_n\right) \leq \sum_{n \in \mathbb{N}}\mu^{*}(A_n)\).
In other words, an outer measure \(\mu^*:\mathcal P(X) \to [0,\infty]\) is a monotone and countably subadditive set function such that \(\mu^*(\varnothing)=0\).
The most common way to obtain outer measures is to start with a family \(\mathcal E\) of “elementary sets” on which a notion of measure is defined (such as rectangles in the plane) and then to approximate arbitrary sets “from the outside” by countable unions of members of \(\mathcal E\).
Proposition. Let \(\mathcal{E} \subseteq \mathcal{P}(X)\) and \(\rho:\mathcal{E} \to [0,\infty]\) be a set function such that \(\varnothing,X \in \mathcal{E}\) and \(\rho(\varnothing)=0\). For any \(A \subseteq X\), define \[\mu^{*}(A)=\inf\bigg\{\sum_{j=1}^{\infty}\rho(E_j):(E_j)_{j\in\mathbb N} \subseteq \mathcal{E} \text{ and } A \subseteq \bigcup_{j=1}^{\infty}E_j\bigg\}.\] Then \(\mu^{*}\) is an outer measure.
Proof. For any \(A \subseteq X\) there exists \((E_j)_{j=1}^{\infty}\) such that \(A \subseteq \bigcup_{j=1}^{\infty}E_j\) (take \(E_j=X\) for all \(j\in\mathbb N\)) so the definition of \(\mu^{*}\) makes sense.
Obviously \(\mu^{*}(\varnothing)=0\) (take \(E_j=\varnothing\) for all \(j\in\mathbb N\)).
We also have \(\mu^{*}(A) \leq \mu^{*}(B)\) for any \(A \subseteq B\). Indeed, if \((E_j)_{j\in\mathbb N} \subseteq \mathcal{E}\) and \(B \subseteq \bigcup_{j=1}^{\infty}E_j\) then \(A \subseteq \bigcup_{j=1}^{\infty}E_j\), so a covering \((E_j)_{j\in\mathbb N} \subseteq \mathcal{E}\) for \(B\) is also a covering for \(A\). Hence \(\mu^{*}(A) \leq \mu^{*}(B)\).
To prove the countable subadditivity, let \((A_j)_{j=1}^{\infty}\subseteq \mathcal P(X)\) and \(\varepsilon>0\). For each \(A_j\) there is \((E_{j, k})_{k=1}^{\infty}\) such that \[A_j \subseteq \bigcup_{k=1}^{\infty}E_{j, k} \quad \text{ and }\quad {\color{blue}\sum_{k=1}^{\infty}\rho(E_{j, k}) \leq \mu^{*}(A_j)+\varepsilon{2^{-j}}}.\]
But then if \(A=\bigcup_{j=1}^{\infty}A_j\) we have \(A\subseteq \bigcup_{j=1}^{\infty}\bigcup_{k=1}^{\infty}E_{j, k}\) and \[\mu^*(A)\le \sum_{j=1}^{\infty}\sum_{k=1}^{\infty}\rho(E_{j, k}). \qquad {\color{red}\text{Why? Justify this!}}\]
Consequently \[\mu^*(A)\le \sum_{j=1}^{\infty}\sum_{k=1}^{\infty}\rho(E_{j, k})\leq \sum_{j=1}^{\infty}{\color{blue}\big(\mu^{*}(A_j)+\varepsilon{2^{-j}}\big)}\le \sum_{j=1}^{\infty}\mu^{*}(A_j)+\varepsilon.\]
Since \(\varepsilon\) is arbitrary, we are done.
Recall from the previous lecture that the family of all rectangles is given by \[\begin{align*} J^d=J_o^d \cup J_{oc}^d \cup J_{co}^d \cup J^d_c, \end{align*}\] where \(J_o^d, J_{oc}^d, J_{co}^d, J^d_c\), are, respectively, the families of open, open-closed, closed-open, and closed rectangles in \(\mathbb R^d\).
Volume function on \(\mathbb R^d\).
For \(I \in J^d\) we define the volume function \({\rm v}_d:J^d \to [0,\infty]\) by \[{\rm v}_d(I)={\rm vol}_d(I)=\begin{cases} 0&\text{ if } I=\varnothing,\\ \prod_{j=1}^{d}(b_j-a_j) &\text{ if } {\rm cl}(I)=\prod_{j=1}^{d}[a_j, b_j],\\ \infty &\text{ otherwise}. \end{cases}\] For \(d=1\) the volume function \({\rm v}_d\) coincides with the length function on \(\mathbb R\).
Problem 1. For every \(E \subseteq \mathbb{R}^d\) we define \[\lambda_{d, o}^{*}(E)=\inf\Big\{\sum_{n \in \mathbb{N}}{\rm v}_d(I_n): (I_n)_{n\in\mathbb N} \subseteq J_{o}^d \text{ and }E \subseteq \bigcup_{n \in \mathbb{N}}I_n\Big\},\] \[\lambda_{d, oc}^{*}(E)=\inf\Big\{\sum_{n \in \mathbb{N}}{\rm v}_d(I_n): (I_n)_{n\in\mathbb N} \subseteq J_{oc}^d \text{ and }E \subseteq \bigcup_{n \in \mathbb{N}}I_n\Big\},\] \[\lambda_{d, co}^{*}(E)=\inf\Big\{\sum_{n \in \mathbb{N}}{\rm v}_d(I_n): (I_n)_{n\in\mathbb N} \subseteq J_{co}^d \text{ and }E \subseteq \bigcup_{n \in \mathbb{N}}I_n\Big\},\] \[\lambda_{d, c}^{*}(E)=\inf\Big\{\sum_{n \in \mathbb{N}}{\rm v}_d(I_n): (I_n)_{n\in\mathbb N} \subseteq J_{c}^d \text{ and }E \subseteq \bigcup_{n \in \mathbb{N}}I_n\Big\}.\] Prove that each of these set functions is an outer measure on \(\mathbb{R}^d\) and for any \(E \subseteq \mathbb{R}^d\) we have \(\lambda_{d, o}^{*}(E)=\lambda_{d, co}^{*}(E)=\lambda_{d, oc}^{*}(E)=\lambda_{d, c}^{*}(E).\)
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 \[\begin{align*} \lambda_d^{*}(E)=\inf\Big\{\sum_{n \in \mathbb{N}}{\rm v}_d(E_n): (E_n)_{n \in \mathbb{N}}\subseteq J_{co}^d\quad \text{ and } \quad E \subseteq \bigcup_{n \in \mathbb{N}}E_n\Big\}. \end{align*}\]
In other words, we have \(\lambda_d^{*}(E)=\lambda_{d, co}^{*}(E)\) for any \(E \subseteq \mathbb{R}^d\).
If \(d=1\) we will abbreviate \(\lambda_d^{*}\) to \(\lambda^{*}\).
Problem 2. For every rectangle \(E\in J^d=J_o^d \cup J_{oc}^d \cup J_{co}^d \cup J_c^d\) prove that \[{\rm v}_d(E)=\lambda_{d, o}^{*}(E)=\lambda_{d, co}^{*}(E)=\lambda_{d, oc}^{*}(E)=\lambda_{d, c}^{*}(E)\]
Carathéodory’s condition. If \(\mu^{*}\) is an outer measure on \(X\), a set \(A \subseteq X\) is called \(\mu^{*}\)-measurable if \[\mu^{*}(E)=\mu^{*}(E \cap A)+\mu^{*}(E \cap A^c) \quad \text{ for all } \quad E \subseteq X.\]
Remark.
By subadditivity of \(\mu^*\) we clearly have \[\mu^{*}(E) \leq \mu^{*}(E \cap A)+\mu^{*}(E \cap A^c).\]
To prove that \(A\subseteq X\) is \(\mu^{*}\)-measurable, it suffices to prove the reverse inequality \[\mu^{*}(E) \geq \mu^{*}(E \cap A)+\mu^{*}(E \cap A^c).\] for all \(E\subseteq X\) such that \(\mu^{*}(E)<\infty\).
Suppose that \(A\subseteq X=(a, b)\times(c, d)\subset\mathbb R^2\) and let \(\mathcal C\) be a family of open rectangles in \(X\) and \(\rho\) be the area function. Define \[\mu^{*}(A)=\inf\bigg\{\sum_{j=1}^{\infty}\rho(C_j):(C_j)_{j\in\mathbb N} \subseteq \mathcal{C} \text{ and } A \subseteq \bigcup_{j=1}^{\infty}C_j\bigg\},\] which is an outer measure that approximates the area of a bounded set \(A\) from the outside by using rectangles.
Since \(\mu^*(X)<\infty\) we can define an inner measure by setting \[\mu_*(A)=\mu^*(X)-\mu^*(A^c),\] which approximates the area of a bounded set \(A\) from the inside.
If \(\mu^{*}(A)=\mu_{*}(A)\) it is reasonable to think that their common value is the area of \(A\), which leads to the measurability condition \[\mu^{*}(A)=\mu_{*}(A)\quad {\color{blue} \Longleftrightarrow \quad \mu^*(X)=\mu^*(X\cap A)+\mu^*(X\cap A^c).}\]
Carathéodory’s Theorem. Let \(\mu^{*}\) be an outer measure on \(X\), and let \[\mathcal M(\mu^*) =\big\{A\subseteq X: \mu^{*}(E)=\mu^{*}(E \cap A)+\mu^{*}(E \cap A^c) \quad \text{ for all } \quad E \subseteq X\big\}.\] be the collection of \(\mu^{*}\)-measurable sets. Then
\(\mathcal M(\mu^*)\) is a \(\sigma\)-algebra, (Carathéodory’s \(\sigma\)-algebra induced by \(\mu^*\)).
\((X, \mathcal M(\mu^*), \mu^{*})\) is a complete measure space.
Proof. Step 1. We first show that \(\mathcal M(\mu^{*})\) is an algebra.
Observe that \(\mathcal M(\mu^{*})\) is closed under complements since the definition of \(\mu^{*}\)-measurability of \(A\) is symmetric in \(A\) and \(A^c\).
Next, if \(A,B \in \mathcal M(\mu^{*})\), then we will show that \(A \cup B \in \mathcal M(\mu^{*}).\)
For any \(E\subseteq X\) we have to show that \[\mu^{*}(E) \geq \mu^{*}(E \cap (A \cup B))+\mu^{*}(E \cap (A \cup B)^c).\]
Note that \((A \cup B)^c=A^c \cap B^c\) and \[A \cup B={\color{purple}(A \cap B)} \cup {\color{purple}(A \cap B^c)} \cup {\color{purple}(A^c \cap B)}.\] Therefore, by subadditivity of \(\mu^*\) and Carathéodory’s condition we obtain \[\begin{gather*} \mu^{*}(E \cap (A \cup B))+\mu^{*}(E \cap (A \cup B)^c)\\ \le {\color{red}\mu^{*}(E \cap A \cap B)+\mu^{*}(E \cap A \cap B^c)}+{\color{blue}\mu^{*}(E \cap A^c \cap B)+\mu^{*}(E \cap A^c \cap B^c)}\\ = {\color{red}\mu^{*}(E \cap A)}+{\color{blue}\mu^{*}(E \cap A^c)} = \mu^{*}(E). \end{gather*}\]
Problem 3. Show that, in fact, we have equality and subadditivity of \(\mu^*\) is not needed!
Finally, we conclude that \[\mu^{*}(E) \geq \mu^{*}(E \cap (A \cup B))+\mu^{*}(E \cap (A \cup B)^c) \geq \mu^{*}(E).\] Thus \(A \cup B \in \mathcal M(\mu^{*})\), which shows that \(\mathcal M(\mu^{*})\) is an algebra.
Step 2. We show that \(\mathcal M(\mu^{*})\) is finitely additive.
Let \(A,B \in \mathcal M(\mu^*)\) be such that \(A \cap B=\varnothing\). Then \({\color{blue}B \subseteq A^c}\) and \[\begin{align*} \mu^{*}(A \cup B)={\color{red}\mu^{*}((A \cup B) \cap A)}+{\color{blue}\mu^{*}((A \cup B) \cap A^c)} ={\color{red}\mu^{*}(A)}+{\color{blue}\mu^{*}(B)}. \end{align*}\]
Step 3. We now show that \(\mathcal M(\mu^{*})\) is a \(\sigma\)-algebra.
Since \(\mathcal M(\mu^{*})\) is an algebra, it suffices to show that \(\mathcal M(\mu^{*})\) is closed under countable disjoint unions. Let \((A_j)_{j=1}^{\infty} \subseteq \mathcal M(\mu^{*})\) be such that \(A_i \cap A_j=\varnothing\) for \(i \neq j\), and define \[B_0=\varnothing, \ \ B_n=\bigcup_{j=1}^{n}A_j, \ \ B=\bigcup_{j=1}^{\infty}A_j.\] Then for any \(E \subseteq X\) we have \[\begin{align*} \mu^{*}(E \cap B_n)&=\mu^{*}(E \cap B_n \cap A_n)+\mu^{*}(E \cap B_n \cap A_n^c)\\ &=\mu^{*}(E \cap A_n)+{\color{red}\mu^{*}(E \cap B_{n-1})}. \end{align*}\]
Iterating \(\mu^{*}(E \cap B_n)=\mu^{*}(E \cap A_n)+{\color{red}\mu^{*}(E \cap B_{n-1})}\) we obtain \[\mu^{*}(E \cap B_n)=\mu^{*}(E \cap A_n)+{\color{red}\mu^{*}(E \cap A_{n-1})+\mu^{*}(E \cap B_{n-2})}=\sum_{j=0}^{n}\mu^{*}(E \cap A_j).\] Therefore we see \[\begin{gather*} \mu^{*}(E)=\mu^{*}(E \cap B_n)+\mu^{*}(E \cap B_n^c)=\sum_{j=0}^{n}\mu^{*}(E \cap A_j)+\mu^{*}(E \cap B_n^c)\\ \geq \sum_{j=0}^{n}\mu^{*}(E \cap A_j)+\mu^{*}(E \cap B^c), \end{gather*}\] since \(B_n\subseteq B=\bigcup_{j=1}^\infty A_j\). Letting \(n \to \infty\) we obtain \[\begin{align*} \mu^{*}(E) \geq \sum_{j=0}^{\infty}\mu^{*}(E \cap A_j)+\mu^{*}(E \cap B^c). \end{align*}\]
Thus by subadditivity of \(\mu^*\) we see that \[\begin{gather*} \mu^{*}(E) \geq \sum_{j=0}^{\infty}\mu^{*}(E \cap A_j)+\mu^{*}(E \cap B^c)\geq \mu^{*}\bigg(\bigcup_{j=0}^{\infty}(E \cap A_j)\bigg) +\mu^{*}(E \cap B^c)\\ =\mu^{*}(E \cap B)+\mu^{*}(E \cap B^c) \geq \mu^{*}(E). \end{gather*}\]
Step 4. We now show that \(\mu^*\) is countably additive on \(\mathcal M(\mu^{*})\).
From the previous calculation we have \[\mu^{*}(E) = \sum_{j=0}^{\infty}\mu^{*}(E \cap A_j)+\mu^{*}(E \cap B^c).\] Taking \({\color{red}E=B=\bigcup_{j=1}^{\infty}A_j}\) we obtain \[\mu^{*}(B)= \sum_{j=1}^{\infty}\mu^{*}( A_j)\] as claimed.
Step 5. We finally show that \(\mu^{*}\) is a complete measure on \(\mathcal M(\mu^{*})\).
Complete measures. Recall that a measure whose domain includes all subsets of null sets is called complete.
Take any \(A \subseteq X\) such that \(\mu^{*}(A)=0\). We will show \(A \in \mathcal M(\mu^{*})\). For any \(E \subseteq X\) we have \[\begin{align*} \mu^{*}(E) \leq \mu^{*}(E \cap A)+\mu^{*}(E \cap A^c) \leq \mu^{*}(E \cap A^c) \leq \mu^{*}(E), \end{align*}\] since \(\mu^{*}(E \cap A)=0\). Thus \[\mu^{*}(E) = \mu^{*}(E \cap A)+\mu^{*}(E \cap A^c)\] and \(A \in \mathcal M(\mu^{*})\). The proof of Carathéodory’s theorem is finished.
Problem 4. Let \((X, \mathcal M, \mu)\) be a measure space. For every \(E\subseteq X\) define \[\mu^*(E)=\inf\{\mu(A): A\supseteq E \text{ and } A\in\mathcal M\}.\]
(a) Prove that \(\mu^*\) is an outer measure on \(X\).
(b) Prove that \(\mathcal M\subseteq \mathcal M(\mu^*)\).
(c) Prove that \(\mu^*(E)=\mu(E)\) for any \(E\in\mathcal M\).
Remark. The purpose of this problem is show that the measure space \((X, \mathcal M(\mu^*), \mu^*)\) may serve as a complete extension of \((X, \mathcal M, \mu)\).
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 \(\mathbb R^d\).
If \(d=1\) we shall abbreviate \(\lambda_d^*\) to \(\lambda^*\) and \(\lambda_d\) to \(\lambda\).
Theorem. One has that \({\rm Bor}(\mathbb R^d)\subseteq \mathcal{L}(\mathbb R^d)\), since \[{\rm Bor}(\mathbb R^d)=\sigma\Big(\Big\{\bigcap_{i=1}^dH_{i, \xi_i}: \xi_1,\ldots, \xi_d\in\mathbb Q\Big\}\Big),\] where \(H_{i, \xi} =\mathbb R^{i-1} \times (-\infty, \xi) \times \mathbb R^{d-i}\) for \(i \in \{1, \ldots, d\}\).
The proof follows from the following proposition.
Proposition. For any \(1\in i\le d\) and \(\xi \in \mathbb R\) the half-space \(H_{i, \xi}\) is Lebesgue measurable.
For any \(a=(a_1, \ldots, a_d)\), \(b=(b_1, \ldots, b_d)\in [-\infty, \infty]^d\) we will write \[[a, b)=\prod_{i=1}^d[a_i, b_i)\in J_{co}^d.\]
We write \(H=H_{i, \xi}\), and we recall that \(\lambda_d^{*}=\lambda_{d, co}^{*}\).
Step 1. We first show that \[{\rm v}_d(I \cap H)+{\rm v}_d(I \setminus H)={\rm v}_d(I) \quad \text{ for every } \quad I \in J_{co}^d.\]
If \(I \subseteq H\) or \(I \cap H=\varnothing\), this is obvious, since in the former case \(I \cap H=I\) and \(I \setminus H=\varnothing\), while in the latter case \(I \cap H=\varnothing\) and \(I \setminus H=I\).
Otherwise \(I=[a,b)\) with \(a_i<\xi<b_i\), and then \(I \cap H=[a, x)\) and \(I \setminus H=[y, b)\), where \[x=(b_1, \ldots, b_{i-1}, \xi, b_{i+1}, \ldots, b_d), \quad y=(a_1, \ldots, a_{i-1}, \xi, a_{i+1}, \ldots, a_d).\]
In particular \(I \cap H\), \(I \setminus H \in J_{co}^d\) and we have \[\begin{align*} {\rm v}_d(I \cap H)+{\rm v}_d(I \setminus H)&=(\xi-a_i)\prod_{j \neq i}(b_j-a_j) +(b_i-\xi)\prod_{j \neq i}(b_j-a_j)\\ &=(b_i-a_i)\prod_{j \neq i}(b_j-a_j)={\rm v}_d(I). \end{align*}\]
Step 2. Let \(A \subseteq \mathbb R^d\) be such that \(\lambda_d^{*}(A)<\infty\) and let \(\varepsilon>0\).
By the definition of \(\lambda_d^{*}\) there is a sequence \((I_j)_{j \in \mathbb N} \subseteq J_{co}^d\) such that \[A \subseteq \bigcup_{j \in \mathbb N}I_j \quad \text{ and } \quad \sum_{j \in \mathbb N}{\rm v}_d(I_j) \leq \lambda_d^{*}(A)+\varepsilon.\]
By Step 1, we have \((I_j \cap H)_{j \in \mathbb N}, (I_j \setminus H)_{j \in \mathbb N} \subseteq J_{co}^d\) and \[A \cap H \subseteq \bigcup_{j \in \mathbb N}(I_j \cap H), \quad A \setminus H \subseteq \bigcup_{j \in \mathbb N}(I_j \setminus H).\]
Hence, by the definition of \(\lambda_d^{*}\) and by Step 1, \[\begin{align*} \lambda_d^{*}(A \cap H)+\lambda_d^{*}(A \setminus H) &\leq \sum_{j \in \mathbb N}{\rm v}_d(I_j \cap H)+\sum_{j \in \mathbb N}{\rm v}_d(I_j \setminus H)\\ &=\sum_{j \in \mathbb N}{\rm v}_d(I_j) \leq \lambda_d^{*}(A)+\varepsilon. \end{align*}\]
Since \(\varepsilon>0\) was arbitrary, \(\lambda_d^{*}(A \cap H)+\lambda_d^{*}(A \cap H^c) \leq \lambda_d^{*}(A)\), and hence \(H\) is \(\lambda_d^{*}\)-measurable as desired.
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 \(\lambda\).
Let \((X, \rho)\) be a metric space. An outer measure \(\mu^*\) is called a metric outer measure if \[\mu^*(A\cup B)=\mu^*(A)+\mu^*(B) \quad\text{ whenever } \quad \rho(A, B)>0.\]
Problem 5. Let \(\mu^*\) be an outer measure on a metric space \((X, \rho)\). Prove that \(\mu^*\) is a metric outer measure if and only if \({\rm Bor}(X)\subseteq \mathcal M(\mu^*)\).
Problem 6. Prove that the Lebesgue outer measure \(\lambda_d^{*}\) is a metric outer measure.
Suppose that \(\mu\) is a finite Borel measure on \(\mathbb R\) and let \(F(x)=\mu((-\infty, x])\) be the distribution function of \(\mu\). Note that
\(F\) 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 \(\mathbb R\).
Problem 7. 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
\[\begin{align*} \mu_{F}^{*}(E)=\inf\Big\{\sum_{n \in \mathbb{N}}(F(b_n)-F(a_n)): E \subseteq \bigcup_{n \in \mathbb{N}}(a_n, b_n]\Big\}. \end{align*}\] Prove that \(\mu_{F}^{*}\) is an outer measure.
Definition. The previous problem and Carathéodory’s theorem immediately yield:
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.
Problem 8. Prove that \(\mu_F^{*}\) is a metric outer measure and deduce \({\rm Bor}(\mathbb R)\subseteq \mathcal{L}_{\mu_F}(\mathbb R)\).
Problem 9. Prove that \(\mu_F((a, b])=F(b)-F(a)\) for any \(-\infty\le a\le b\le \infty\).
Problem 10. Prove that the Lebesgue–Stieltjes measure \(\mu_F\) is unique in the sense that if \(G\) is another such function, we have \(\mu_F=\mu_G\iff F-G\) 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 \(\mu=\mu_F\).
Definition. Let \((X, \rho)\) be a metric space, let \(p \geq 0\) and \(\delta>0\). For every \(A \subseteq X\) we define
\[H_{p, \delta}(A)=\inf\Big\{\sum_{j \in \mathbb N}({\rm diam}\, B_j)^{p}: A \subseteq \bigcup_{j \in \mathbb N}B_j \ \text{ and } \ {\rm diam}\, B_j \leq \delta\Big\},\] with the convention that \(\inf \varnothing=\infty\). The \(p\)-dimensional Hausdorff outer measure of \(A\) is
\[H_p(A)=\lim_{\delta \to 0^{+}}H_{p, \delta}(A).\]
Here \({\rm diam}\, B=\sup\{\rho(x,y): x, y \in B\}\), and we use the conventions \({\rm diam}\, \varnothing=0\) and \(({\rm diam}\, B)^{0}=1\) for \(B \neq\varnothing\).
As \(\delta\) decreases, the infimum is taken over a smaller family of coverings of \(A\), so \(H_{p, \delta}(A)\) increases and the above limit always exists in \([0, \infty]\).
Problem 11. Let \((X, \rho)\) be a metric space and let \(p \geq 0\). Prove that
(a) \(H_{p, \delta}\) is an outer measure on \(X\) for every \(\delta>0\);
(b) \(H_{p, \delta}(A) \leq H_{p, \delta'}(A)\) for every \(A \subseteq X\) and \(0<\delta'\leq \delta\), and consequently \[H_p(A)=\lim_{\delta \to 0^{+}}H_{p, \delta}(A)=\sup_{\delta>0}H_{p, \delta}(A);\]
(c) \(H_p\) is an outer measure on \(X\).
Problem 12. Prove that the value of \(H_{p, \delta}(A)\) does not change if the sets \(B_j\) are required to be closed, and also if they are required to be open. If \(X=\mathbb R\), prove that one may equally well restrict the \(B_j\) to be closed intervals or open intervals.
Hint to Problem 12: for closed sets it suffices to observe that \({\rm diam}\, B_j={\rm diam}\, {\rm cl}(B_j)\); for open sets one replaces \(B_j\) by \[U_j=\{x \in X: \rho(x, B_j)<\varepsilon 2^{-j-1}\},\] whose diameter is at most \(({\rm diam}\, B_j)+\varepsilon 2^{-j}\).
When \(p\) is an integer, the intuition behind the definition of \(H_p\) is that if \(A\subseteq \mathbb R^d\) behaves like a \(p\)-dimensional set — for instance, if \(A\) is a relatively open subset of a \(p\)-dimensional linear subspace — then the portion of \(A\) lying in a region of diameter \(r\) should have \(p\)-dimensional size comparable to \(r^p\).
Requiring the covering sets to have arbitrarily small diameters is essential for \(H_p\) to capture the local geometry of irregular sets. Without this restriction, one could cover \(A\) simply by \(A\) itself, obtaining the crude bound \[H_p(A)\leq (\operatorname{diam} A)^p,\] which generally does not reflect the true \(p\)-dimensional size of \(A\).
Problem 13. Prove that \(H_p\) is a metric outer measure on \((X, \rho)\) and deduce from Problem 5 that \({\rm Bor}(X) \subseteq \mathcal M(H_p)\).
Definition. The restriction of \(H_p\) to \({\rm Bor}(X)\) is a measure, which we still denote by \(H_p\) and call the \(p\)-dimensional Hausdorff measure.
Problem 14. Prove that \(H_p\) is invariant under isometries of \((X, \rho)\). Moreover, if \(Y\) is a set and \(f, g: Y \to X\) satisfy \[\rho(f(y), f(z)) \leq C\rho(g(y), g(z)) \quad \text{ for all } \quad y, z \in Y,\] prove that \(H_p(f(A)) \leq C^{p}H_p(g(A))\) for every \(A \subseteq Y\).
Problem 15. Let \(A \subseteq X\). Prove that
(a) if \(H_p(A)<\infty\), then \(H_q(A)=0\) for every \(q>p\);
(b) if \(H_p(A)>0\), then \(H_q(A)=\infty\) for every \(q<p\).
Definition. By Problem 15, for every \(A \subseteq X\) the numbers \[\inf\{p \geq 0: H_p(A)=0\} \quad \text{ and } \quad \sup\{p \geq 0: H_p(A)=\infty\}\] are equal. Their common value is called the Hausdorff dimension of \(A\) and is denoted by \({\rm dim}_H(A)\).
Problem 16. If \(X=\mathbb R^d\) prove that there is a constant \(\gamma_d>0\) such that \(\gamma_d H_d\) is the Lebesgue measure on \(\mathbb R^d\). In particular, \({\rm dim}_H(\mathbb R^d)=d\).