Extension Theorem applied to Stieltjes Measures of Real Line.

Some of this presentation borrows liberally from Durrett’s book.

definition: A system of sets of the set $\Omega$ is a collection of subsets of $\Omega$ .

definition: A $\pi$ -system of $\Omega$ is a system of sets of $\Omega$ which is closed under intersection.

definition: a semi-algebra of $\Omega$ is a system of sets $S$ of $\Omega$ that is closed under intersection and such that the complement of any set in the system is a finite disjoint union of sets of the system.

definition: an algebra of $\Omega$ is a system of sets $A$ that is closed under intersection and complement.

lemma: the system of finite unions of sets $A$ of a semi-algebra $S$ is the smallest algebra containing $S$ .

proof: Suppose that $b=\cup_{i=1\ldots,n} a_i$ is a finite disjoint union of sets of $S$ . Then for each $a_i$ the complement $a_i^C$ is a finite disjoint union of sets $a_{ij}$ of $S$ .

$b^C = \cap_{i=1,\ldots,n} a_i^C = \cap_{i=1..n} \cup_{j=1..n_i} a_{ij} = \cup_{\pi \in \prod_{i=1..n} 1..n_i} \cap_{i = 1..n} a_{i,\pi(i)}$ . The intersection of sets of $S$ is a set of $S$ . Let $\pi,\pi' \in \prod_{i=1..n} 1..n_i$ . Then if

$\cap_{i = 1..n} a_{i,\pi(i)} \cap \cap_{i = 1..n} a_{i,\pi'(i)} = \cap_{i = 1..n} a_{i,\pi(i)} \cap a_{i,\pi'(i)} \neq \emptyset$

it follows that for all $i = 1..n$ , $\pi(i) = \pi'(i)$ since $\{ a_{ij} : j = 1,\ldots,n(i)\}$ is a disjoint collection. Hence the complement is a disjoint union of sets of $S$ and so is an element of $A$ .

If $b,c\in A$ then there are disjoint $b_1,\ldots,b_m$ and disjoint $c_1,\ldots,c_n$ such that $b = \cup_i b_i$ and $c = \cup_j c_j$ . The intersection

$b \cap c = \left( \cup_{i=1..m} b_i \right) \cap \left( \cup_{j=1..n} c_j \right) = \left( \cup_{i=1..m} b_i \cap \left( \cup_{j=1..n} c_j \right) \right) = \cup_i,j b_i \cap c_j$

and since the intersection of elements of $S$ is an element of $S$ , and if $b_i \cap c_j \cap (b_k \cap c_l) \neq \emptyset$ implies $b_i \cap b_k \neq \emptyset$ and $c_j \cap c_l \neq \emptyset$ and so $i = k, j = l$ . Hence the intersection of two elements of $A$ is a finite disjoint union of elements of $S$ .

Hence the set $A$ is closed under complement and finite intersection and so is an algebra.

Let $G$ be any algebra containing the semi-algebra $S$ . Let $a$ be a finite union of elements of $S$ . Then since $G$ is closed under finite union, it contains $a$ . Since this is true for any such $a$ , it $G$ contains $A$ .

lemma: Let $S$ be a semialgebra of $\Omega$ and $\mu$ be a real-valued function on $S$ such that $\mu(\emptyset) = 0$ and when $\cup_{i=1..n} a_i$ is an element of $S$ and $a_1,\ldots,a_n$ are disjoint elements of $S$ , $\mu(\cup_{i=1..n} a_i) = \sum_{i=1..n} \mu a_i$ . Then

when $A = a(S)$ is the algebra generated from $S$ , then

for any $a \in A$ that is the finite disjoint union of sets $a_1,\ldots,a_n$ of $A$ then $\mu(\cup_{i = 1..n} a_i) = \sum_{i = 1..n} \mu(a_i)$ .
for any $a \in A$ that is contained in a finite union of sets $a_1,\ldots,a_n$ of $A$ then $\mu(a) \le \sum_{i=1..n} \mu(a_i)$ .

proof:

In the first case, since each $a_i$ is also a finite disjoint union of sets $a_{ij},j=1..n_i$ where $a_{ij}$ is an element of $S$ , it follows that their union is a finite disjoint union of sets of $S$ . And so

$\mu(\cup_{i=1..n} a_i) = \sum_{i =1..n} \sum_{j = 1..n_i} \mu(a_{ij}) = \sum_{i = 1..n} \mu(a_i)$ .

The second case is harder to establish. A proof by induction is a good starting point. If $a \subseteq b$ for elements $a,b \in a(S)$ , it follows that $b = a \cup (b \cap a^C)$ , is disjoint and the first component is a finite disjoint union of elements of $S$ , and the second component is also a finite disjoint union of elements of $S$ , so the (disjoint) union of the two components is a disjoint union of elements of $S$ .

Assume the claim is valid for all sets of $A$ that are contained in a union of at most $n-1$ elements of $A$ . Then

$\cup_{i = 1..n} b_i = \cup_{i = 1..n} \left( b_i - \cup_{j < i} b_j \right)$

is a disjoint finite union of sets of $a(S)$ . and since

$a = \cup_{i = 1..n} a \cap b_i = \cup_{i = 1..n} a \cap \left[ \left( b_i - \cup_{j < i} b_j \right) \right]$

is a disjoint union of sets of $a(S)$ , and by the first case

$\mu(a) = \sum_{i=1..n} \mu(a \cap \left( b_i - \cup_{j < i} b_j \right) )$

but since the case $n = 1$ has been established,

$\mu(a \cap \left( b_i - \cup_{j < i} b_j \right) ) \le \mu( \left( b_i - \cup_{j < i} b_j \right))$ and so

$\mu(a) \le \sum_{i = 1..n} \mu( \left( b_i - \cup_{j < i} b_j \right)) = \mu(b)$

theorem: Let $S$ be the semi-algebra of open-closed intervals of the extended real line, and let $F$ be a stieltjes function of the real line. Define $\mu( (a,b] ) = F(b) - F(a)$ . Then there is a unique measure the borel sets of the real line that is consistent with $\mu$ .

proof: the proof utilizes the extension theorem and the prior lemma.

If $(a,b] = \cup_{i=1..n} (a_i,b_i]$ is a disjoint union, wlog one may assume that the intervals are ordered by increasing left endpoint, in which case $a = a_1$ and $b = b_n$ , and $b_i = a_{i+1}$ for $i = 1\ldots,n-1$ . so

$\sum_{i = 1..n} \mu((a_i,b_i]) = \sum_{i=1..n} F(b_i) - F(a_i) = \sum_{i=1..n-1} F(a_{i+1}) - F(a_i) + F(b) - F(a_n) = F(a_n) - F(a) + F(b) - F(a_n) = F(b) - F(a) = \mu((a,b])$

establishing the first requirement of the extension theorem.

Assume $a < b$ are real numbers. If $(a,b] \subseteq \cup_{i=1..n} (a_i,b_i]$ then wlog assume $a_i < b_i$ are real numbers.

Since $F$ is right-continuous, for any $\epsilon > 0$ there is a $\delta > 0$ such that $F(a + \delta) < F(a) + \epsilon$ and for each $i$ there is a $\nu_i > 0$ such that $F(b_i + \nu_i) < F(b_i) + \epsilon 2^{-i}$ .

Since the open intervals $(a_i, b_i + \nu_i)$ cover $[a + \delta, b]$ and by the Heine Borel theorem the open intervals have a finite subcover. Hence by the second conclusion of the support lemma,

$\mu((a+\delta,b]) \le \sum_{i \in I} \mu( (a_i, b_i + \nu_i]) \le \sum_i \mu( (a_i, b_i + \nu_i])$

and so

$F(b) - F(a) - \epsilon < F(b) - F(a + \delta) \le \sum_i F(b_i + \nu_i) - F(a_i) \le \epsilon + \sum_i F(b_i) - F(a_i)$

a statement true for all $\epsilon > 0$ . hence the second requirement of the extension theorem is valid when $a,b$ are real numbers such that $a < b$ .

If $(a,b] \subseteq \cup_i (a_i,b_i]$ when $a,b$ are extended real numbers, then for any $A,B$ that are real numbers such that $a < A < B < b$

$(A,B] \subseteq \cup_i (a_i,b_i]$ and so

$F(B) - F(A) \le \sum_i F(b_i) - F(a_i)$ .

Since this is valid for all such $A,B$ , it follows that

$F(b) - F(a) \le \sum_i F(b_i) - F(a_i)$

and the second condition of the extension theorem is satisfied. Hence the measure can be extended to the algebra containing the semi-algebra of open-closed intervals, and since the stieltjes function is bounded, it is sigma finite and therefore the measure can be extended to the sigma algebra containing the semi-algebra. But this collection coincides with the borel sets.

Extension Theorem applied to Stieltjes Measures of Real Line.

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