Lebesgue Measure Construction

Measure Space

Let $\Omega$ be a set and $2^\Omega$ be the power set of $\Omega$.

$\sigma$-Algebra

Definition. A collection of subsets $\mathcal{F} \subseteq 2^\Omega$ is a $\sigma$-algebra (or $\sigma$-field) if

• It is closed under countable union, intersection and complement.

Measure

Definition. Given a collection of subsets $\mathcal{F} \subseteq 2^\Omega$, a function $\mu : \mathcal{F} \rightarrow \mathbb{R}$ is a measure if it satisfies

1. It is non-negative: $\forall A \in \mathcal{F}, \mu(A) \ge 0$.

2. The empty set is mapped to 0: $\mu(\emptyset) = 0$.

3. $\sigma$-additivity: if $A_i \in \mathcal{F}$ is a countable sequence of disjoint sets, s.t., $\cup_{i \ge 1} A_i \in \mathcal{F}$, then

$$\mu(\cup_{i \ge 1} A_i) = \sum_{i \ge 1} \mu(A_i ).$$

Measure Space

Definition. A measure space $(\Omega, \mathcal{F}, \mu)$ consists of three part:

1. A set $\Omega$, on which it is defined;

2. A $\sigma$-field $\mathcal{F}$ that contains all measurable sets;

3. A measure $\mu$ defined on $\mathcal{F}$.

In general, it is not easy to construct a measure space on $\Omega$ directly. Our indirect construction consists of three steps:

1. Consider a simple set class $S$ and a measure $\mu$ defined on $S$.
2. We extend the domain of $\mu$ from $S$ to a larger one $\Omega=2^\Omega$.
3. We choose a subset from $2^\Omega$, such that it is a $\sigma$-field and $\mu^*$ is a measure defined on it.

In what follows, we illustrate the idea via constructing measures on $\mathbb{R}^d$.

Semialgebra

Definition. A class of subsets $S \subseteq 2^\Omega$ is a semialgebra if

1. It is closed under intersection: $\forall A, B \in S$, $A \cap B \in S$.

2. If $A \in S$, then $\bar A$ is a finite disjoint union of sets in $S$.

Examples. For $d \in \mathbb{Z}^+$, the collection of half-open intervals in $\mathbb{R}^d$ plus the empty set

$$S_d \doteq \{ \times_{i \in [d] } (a_i, b_i] : -\infty \le a_i \lt b_i \le \infty, \forall i \in [d] \} \cup \{ \emptyset \}$$

Algebra

Definition. A class of subsets $\mathcal{F} \subseteq 2^\Omega$ is an algebra (or field) if

• It is closed under union, intersection and complement.

Remark. By De Morgan's laws, $\overline{A \cap B } = \bar A \cup \bar B$ and $\overline{A \cup B} = \bar A \cap \bar B$. If we know a collection is closed under intersection (union) and complement, so is union (intersection).

Theorem If $S \subseteq \Omega$ is a semialgebra, then

$$\widetilde{S} \doteq \{ \text{finite disjoint union of sets in } S \}$$

is an algebra.

Proof.

1. $\widetilde{S}$ is closed under intersection: Let $A, B \in \widetilde{S}$ with $A = +_i A_i$ and $B = +_j B_j$, where $+$ denotes disjoint union and $A_i, B_j \in S$. Then $A \cap B = +_{i, j} A_i \cap B_j$. As $S$ is a semialgebra, it follows that $A_i \cap B_j \in S$ and $A \cap B \in \widetilde{S}$.

2. $\widetilde{S}$ is closed under complement. Let $A \in \widetilde{S}$ with $A = +_i A_i$, where $A_i \in S$. Then $\bar A = \cap_i \bar A_i$. As $S$ is a semialgebra, $\bar A_i$ is a finite disjoint union of sets in $S$, which implies that $\bar A_i \in \widetilde{S}$. Therefore $\bar A$ is a finite intersection of sets in $\widetilde{S}$. Since $\widetilde{S}$ is closed under intersection, $\bar A \in \widetilde{S}$.

$\square$

Measure Extension: Semialgebra $\rightarrow$ Algebra

Theorem. Let $S$ be a semialgebra and $\mu$ be a measure on $S$. Then $\mu$ has a unique extension $\bar \mu$ that is a measure on $\widetilde S$, the algebra generated by $S$.

Remark. The $\sigma$-additivity condition of $\mu$ being a measure can be replaced by

1. Finite additivity. If $A, A_i \in S$ with $A = +_i A_i$, then $\mu(A) = \sum_i \mu (A_i)$.
2. Subadditivity. If $A, A_i \in S$ with $A = +_{i \ge 1} A_i$, then $\mu (A) \le \sum_{i \ge 1} \mu (A_i)$.

The subscript $i \ge 1$ in the conditions indicates a countable union, while a plain subscript $i$ or $j$ indicates a finite union.

The relation between the condition stated in the Theorem and the relaxed ones is established as follows.

Exercise. A finitely additive nonnegative set function on an algebra $\widetilde{S}$ is monotone and the following conditions are equivalent

1. $\sigma$-additivity.
2. Subadditivity.

The proof for the Theorem consists of two steps.

$\blacktriangleright$ Defining $\bar \mu$

If $\mu$ is finitely additive in $S$, then it has a finitely additive extension $\bar \mu$ in $\widetilde{S}$.

1. Extension. If $A = +_i A_i \in \widetilde{S}$ with $A_i \in S$, define

   $$\bar \mu(A) = \sum_i \mu (A_i).$$

   We need to verify the value of $\bar \mu(A)$ is unique. Suppose $A$ has another decomposition: $A = +_j B_j$ with $B_j \in S$. Since $S$ is a semialgebra, $A_i \cap B_j \in S$. Using that $\mu$ is a measure

   $$\sum_i \mu (A_i) = \sum_{i, j} \mu(A_i \cap B_j) = \sum_j \mu(B_j).$$

   It follows that $\bar \mu(A)$ does not change with the decomposition of $A$ and hence is well-defined.

2. Finite additivity. Let $A = +_i A_i \in \widetilde{S}$ with $A_i \in \widetilde{S}$ and suppose that $A_i = +_j B_{i,j}$ with $B_j \in S$. Then $A = +_{i,j} B_{i,j}$. By definition of $\bar \mu (A)$ and $\bar \mu (A_i)$,

We list two more properties about $\bar \mu$ induced by finite additivity.

• Monotonicity. If $A, B \in \widetilde{S}$ with $A \subseteq B$, then $\bar \mu (A) \le \bar \mu (B)$. As $\widetilde{S}$ is an algebra, $B - A \in \widetilde{S}$ and by finite additivity of $\bar \mu$,

  $$\bar \mu (A) \le \bar \mu (A) + \bar \mu (B - A) = \bar \mu (B).$$

• Finite subadditivity. If $A \subseteq \cup_{i \in [n] } A_i$ with $A \in \widetilde{S}$, $A_i \in \widetilde{S}, \forall i \in [n]$, then $\bar \mu(A) \le \sum_i \bar \mu (A_i)$. Let $B_1 = A_1$ and $B_i = A_i - A_{i - 1} - \ldots - A_1$ for $i \ge 1$. As $\widetilde{S}$ is an algebra, $B_i \in \widetilde{S}$. By monotonicity and finite additivity of $\bar \mu$,

  $$\bar \mu (A) \le \bar \mu( +_i B_i ) = \sum_i \bar \mu (B_i) \le \sum_i \bar \mu (A_i).$$

$\square$

$\blacktriangleright$ $\sigma$-Additivity of $\bar \mu$

If $A, A_1, \ldots \in \widetilde S$ with $A = +_{i \ge 1} A_i$, then

$$\small \bar \mu (A) = \sum_{i \ge 1} \bar \mu (A_i).$$

Proof. Let $A_i = +_j B_{i,j}$ with $B_{i,j} \in S$. Then $A = +_{i,j} B_{i,j}$ and $\sum_{i \ge 1} \bar \mu(A_i) = \sum_{i \ge 1} \sum_j \mu ( B_{i,j} )$. WLOG, we assume that $A_i$ are countable union of disjoint sets in $S$.

Next, assume that $A = +_k C_k$ with $C_k \in S$. Using that $S$ is closed under intersection, and that $\mu$ is a measure on $S$ (therefore $\sigma$-additivity holds),

Observing that nonnegative numbers can be summed in any order,

$\square$

Measure Extension: Algebra $\rightarrow$ $\sigma$-Algebra

Definition. $\bar \mu$ is said to be $\sigma$-finite if there is a sequence of sets $A_i \in \widetilde{S}$ such that $\mu(A_i) < \infty$ and $\cup_{i \ge 1} A_i = \Omega$.

Caratheodory’s Extension Theorem. Let $\bar \mu$ be a $\sigma$-finite measure on an algebra $\widetilde S$. Then $\bar \mu$ has a unique extension $\mu^*$ (that is a measure) to $\sigma(\widetilde S)$, the smallest $\sigma$-field containing $\widetilde S$.

Remark: Observe that $\sigma(\widetilde S) = \sigma(S)$.

The proof is included at the end of the article.

Lebesgue-Stieltjes Measure

In this section, we construct concrete measures on $\mathbb{R}^d$.

Definition. A function $F: \mathbb{R} \rightarrow \mathbb{R}$ is a Stieltjes measure function if

1. $F$ is non-decreasing.

2. $F$ is right continuous, i.e., $\lim_{y \downarrow x} F(y) = F(x)$.

Definition (Borel Sets). Let $\mathcal{R}^d = \sigma (S_d)$. It is the smallest $\sigma$-field containing all open sets in $\mathbb{R}^d$.

Theorem 3. Associated with each Stieltjes measure function $F$ there is a unique measure $\mu$ on $(\mathbb{R},\mathcal{R}^1)$ with

$$\mu \big( (a, b] \big) = F(b) - F(a).$$

When $F(x) = x$, the measure $\mu$ is called Lebesgue measure.

Proof. We just to verity that $\mu$ is a measure defined on the following semi-algebra:

1. $\mu((a, b]) \ge 0$, since $F$ is non-decreasing.

2. Finite additivity. If $(a, b] = +_i (a_i, b_i]$, $\mu \big( (a, b] \big) = \sum_i \mu \big( ( a_i, b_i ] \big)$. We can rearrange the $\{ (a_i, b_i] \}$, such that $a = a_1$, $b_i = a_{i + 1}$ for all $1 \le i < n$, and $b_n = b$. Hence,

   $$\mu \big( (a, b] \big) = F(b) - F(a) = \sum_i ( F(b_i) - F(a_i) ) = \sum_i \mu \big( ( a_i, b_i ] \big).$$

3. Finite subadditivity. If $(a, b] \subseteq \cup_{i \in [n] } (a_i, b_i]$, $\mu \big( (a, b] \big) \le \sum_{i \in [ n ] } \mu \big( ( a_i, b_i ] \big)$. Recall that the definition of $\bar \mu$ only requires finite additivity of $\mu$. Working with $\widetilde{S_1}$ and $\bar \mu$ significantly simplifies the proof. As for all $A \in S_1$, it holds that $\mu(A) = \bar \mu(A)$. It suffices to prove that

   $$\bar \mu \big( (a, b] \big) \le \sum_{i \in [ n ] } \bar \mu \big( (a_i, b_i] \big).$$

   But this follows exactly from finite subadditivity of $\bar \mu$.

4. $\sigma$-additivity. If $(a, b] = +_{i \ge 1} (a_i, b_i]$, then $\mu \big( (a, b] \big) = \sum_{i \ge 1} \mu \big( ( a_i, b_i ] \big)$. Using finite subadditivity of $\mu$, for all $n \in \mathbb{Z}^+$,

   $$\mu \big( (a, b] \big) \ge \sum_{i \in [n] } \mu \big( ( a_i, b_i ] \big).$$

   Letting $n \rightarrow \infty$, we get $\mu \big( (a, b] \big) \ge \sum_{i \ge 1} \mu \big( ( a_i, b_i ] \big)$.

   It is left to prove the reverse direction of the inequality ($\sigma$-subadditivity). If $\mu \big( ( a_i, b_i ] \big) = \infty$ for some $i$, then the inequality is trivial. WLOG, assume that $-\infty < a_i < b_i < \infty$ for all $i$.

   First, suppose that both $a$ and $b$ are finite. For each $\epsilon > 0$, as $F$ is right continuous, there exists some $\delta > 0$, such that

   $$\begin{aligned} F(a + \delta) &\le F(a) + \epsilon\,,\qquad \text{ and } \\ \mu \Big( (a, b] \Big) &\le \mu \Big( (a + \delta, b] \Big) + \epsilon. \end{aligned}$$

   For each $i$, there exists some $\delta_i > 0$, such that

   $$\begin{aligned} F(b_i + \delta_i) &\le F(b_i) + \epsilon / 2^i\,,\qquad \text{ and } \\ \mu \Big( (a_i, b_i + \delta_i] \Big) &\le \mu \Big( (a_i, b_i] \Big) + \epsilon / 2^i. \end{aligned}$$

   Since $\{ (a_i, b_i + \delta_i) \}$ is an open cover of the closed interval $[a + \delta, b]$, there is a finite sub-cover $\{ (s_i, t_i) \}$. It follows that $\{ (s_i, t_i] \}$ covers $(a + \delta, b]$. Therefore,

   $$\mu \Big( (a + \delta, b] \Big) \le \sum_i \mu \Big( (s_i, t_i] \Big) \le \sum_{i \ge 1} \mu \Big( (a_i, b_i + \delta_i] \Big)$$

   It follows that

   $$\mu \Big( (a, b] \Big) \le \sum_{i \ge 1} \mu \Big( (a_i, b_i] \Big) + 2 \epsilon.$$

   Letting $\epsilon \rightarrow 0$ finishes the proof.

   If $a = -\infty$ or $b = \infty$, the inequality $\small \mu \Big( (a', b'] \Big) \le \sum_{i \ge 1} \mu \Big( (a_i, b_i] \Big)$ holds for all finite subset $(a', b'] \subseteq (a, b]$. Hence it holds for $(a, b]$.

$\square$

Caratheodory’s Extension Theorem

Caratheodory’s Extension Theorem. Let $\bar \mu$ be a $\sigma$-finite measure on an algebra $\widetilde S$. Then $\bar \mu$ has a unique extension $\mu^*$ (that is a measure) to $\sigma(\widetilde S)$, the smallest $\sigma$-field containing $\widetilde S$.

$\spadesuit$ Outer Measure

Define the function $\mu^*: 2^\Omega \rightarrow \mathbb{R}_{\ge 0}$ as $\forall A \subseteq \Omega$,

Intuitively, $\mu^*$ is an upper bound on the measure of $A$. The following properties follows immediately form the definition.

1. $\mu^*(\emptyset) \le \sum_{i = 1}^\infty \bar{ \mu }(\emptyset) = 0$.

2. $\mu^*$ is monotone: $A \subseteq B$ implies that $\mu^*(A) \le \mu^*(B)$.

3. Subadditivity. If $A \subseteq \cup_{i \ge 1} A_i$, then $\mu^*(A) \le \sum_{i \ge 1} \mu^*(A_i)$. Given $\epsilon > 0$, for each $A_i$, there exists a countable sequence $\{ B_{i, j} : j \in \mathbb{Z}^+ \}$ in $\widetilde{S}$ such that it covers $A_i$ and

   $$\sum_{j \ge 1} \bar{ \mu }(B_{i, j}) \le \mu^*(A_i) + \epsilon / 2^i.$$

   The sequence $\{ B_{i, j} : i, j \in \mathbb{Z}^+ \}$ is countable and covers $A$. Further

   $$\mu^*( A ) \le \sum_{i \ge 1} \sum_{j \ge 1} \bar{ \mu }(B_{i, j}) \le \sum_{i \ge 1} \mu^* (A_i) + \epsilon.$$

   The inequality holds for all $\epsilon > 0$ hence it holds that

   $$\mu^*( A ) \le \sum_{i \ge 1} \mu^* (A_i).$$

Now, $\mu^*$ is almost a measure: it is an nonnegative set function and satisfies subadditivity. By previous analysis, we know that

Given a nonnegative, finitely additive set function defined on an algebra, the following conditions are equivalent

• $\sigma$-additivity.
• $\sigma$-subadditivity.

If we can prove that $\mu^*$ is finitely additive, then it is a measure on the $\sigma$-algebra, $2^\Omega$ (which is also an algebra). However, in general, finite additivity does not hold on $2^\Omega$. Therefore, we restrict our discussion on a subset of $2^\Omega$.

Measurable Set

Intuitively, the measurable sets are the ones on which finite additivity of $\mu^*$ holds. Formally, we say that a set $A \subseteq \Omega$ is called measurable if

Using subadditivity, to prove $A$ is measurable, it is enough to show that

Definition. $S^* =$ { the collection of all measurable sets }.

Theorem. $S^*$ is a $\sigma$-field that contains $\widetilde{S}$ and the restriction of $\mu^*$ to $S^*$ is a measure.

$\blacktriangleright \quad S^*$ is a $\sigma$-field

1. If $A \in S^*$, then so is $\bar A$.

   Proof. By definition.

2. If $A_1, A_2 \in S^*$, then $A_1 \cup A_2 \in S^*$ and $A_1 \cap A_2 \in S^*$.

   Proof. Observe that

   $$A_1 \cup A_2 \subseteq (A_1 \cap \bar A_2) \cup (A_1 \cap A_2) \cup (\bar A_1 \cap A_2).$$

   For all $B \subseteq \Omega$, it holds that

   $$\begin{aligned} \mu^*(B) &= \mu^*(B \cap A_1) + \mu^* (B \cap \bar A_1) \\ &= \mu^*(B \cap A_1 \cap A_2 ) + \mu^*(B \cap A_1 \cap \bar A_2 ) \\ &\quad + \mu^* (B \cap \bar A_1 \cap A_2 ) + \mu^* (B \cap \bar A_1 \cap \bar A_2 ) \\ &\ge \mu^*(B \cap (A_1 \cup A_2) ) + + \mu^* (B \cap \bar A_1 \cap \bar A_2 ) \end{aligned}$$

   The last inequality follows form subadditivity of $\mu^*$. Hence $A_1 \cup A_2 \in S^*$. Finally, $A_1 \cap A_2 = \overline{ \bar A_1 \cup \bar A_2} \in S^*$.

   $\square$

3. If $A_i \in S^*$, then $A = \cup_{i \ge 1} A_i \in S^*$.

   Proof. We have proven that $S^*$ is an algebra. Let $B_1 = A_1$ and $B_i = A_i - A_{i - 1} - \ldots - A_1$ for $i \ge 2$. It holds that $B_i \in S^*$ and $\cup_i A_i = \cup B_i$. Replacing $A_i$ with $B_i$ we can assume without lose of generality that the $A_i$ are disjoint. For all $n \in \mathbb{Z}^+$, let $C_n = +_{i \in [n] } A_i$. For all $B \subseteq \Omega$ and $n > 1$,

   $$\begin{array}{ll} \mu^*(B) &= \mu^* (B \cap C_n ) + \mu^* (B \cap \bar C_n ) \\ &= \mu^* (B \cap C_n \cap A_n ) + \mu^* (B \cap C_n \cap \bar A_n ) + \mu^* (B \cap \bar C_n ) \\ &= \mu^* (B \cap A_n ) + \mu^* (B \cap C_{n - 1} ) + \mu^* (B \cap \bar C_n ) \\ \end{array}$$

   The last equality follows from that $C_n \cap A_n = A_n$ and $C_n \cap \bar A_n = C_{n - 1}$. By induction, and monotonicity of $\mu^*$, we get

   $$\begin{array}{ll} \mu^*(B) &= \sum_{i \in [n] } \mu^* (B \cap A_i ) + \mu^* (B \cap \bar C_n ) \\ &\ge \sum_{i \in [n] } \mu^* (B \cap A_i ) + \mu^* (B \cap \bar A ) \end{array}$$

   Letting $n \rightarrow \infty$ finishes the proof.

   $\square$

$\blacktriangleright \quad S^* \supseteq \widetilde{S}$

Proof. For all $B \subseteq \Omega$, and all $\epsilon > 0$, there exists a sequence $\{ B_i \}$ in $\widetilde{S}$, such that $B \subseteq \cup_i B_i$ and

Let $A \in \widetilde{S}$. Observe that $B_i \cap A, B_i \cap \bar A \in \widetilde{S}$, and that

By the definition of $\mu^*$ and finite additivity of $\bar \mu$, it holds that

Letting $\epsilon \rightarrow 0$, it concludes that

$\square$

$\blacktriangleright \quad \mu^*$ is a measure restricted to $S^*$

Sigma additivity of $\mu^*$. Let $A = +_{i \ge 1} A_i$, with $A_i \in S^*$, then

Proof. As $S^*$ is an algebra and $\mu^*$ is $\sigma$-subadditive, it suffices to prove that $\mu^*$ is finitely additive. Let $A = +_{i \in [n] } A_i$.

$\square$

$\spadesuit$ Uniqueness

$\blacktriangleright\,\,\pi$-system

Definition. $\mathcal{P}$ is said to be a $\pi$-system if it is closed under intersection.

$\blacktriangleright\,\,\lambda$-system

Definition. $\mathcal{L}$ is said to be a $\lambda$-system if
(1). $\Omega \in \mathcal{L}$.
(2). If $A \in \mathcal{L}$, then $\bar A \in \mathcal{L}$.
(3). If $A_1, A_2, \ldots \in \mathcal{L}$ with $A_n \cap A_m = \emptyset$ for $n \neq m$, then $\cup_{n \ge 1} A_n \in \mathcal{L}$.

Note that a $\lambda$-system is almost a $\sigma$-algebra, expect that it is only closed under disjoint union.

Lemma. Replacing (2) in the definition with the following one results in equivalent definition
(2'). If $A, B \in \mathcal{L}$ with $A \subseteq B$, then $B - A \in \mathcal{L}$.

Proof.

Assume that (1), (2), (3) hold. If $A, B \in \mathcal{L}$ with $A \subseteq B$, then

Hence (2') holds.

Assume that (1), (2'), (3) hold. Since $\Omega \in \mathcal{L}$, if $A \in \mathcal{L}$, then

Hence (2) holds.

$\square$

Lemma. If a $\lambda$-system $\mathcal{L}$ is closed under intersection, then it is a $\sigma$-field.

Proof. Since it is closed under complement and intersection, it is an algebra. It is a $\sigma$-field as it satisfies monotonicity.

$\square$

$\blacktriangleright\,\,$ $\pi \text{-} \lambda$ Lemma

$\pi \text{-} \lambda$ Lemma. Let $\mathcal{P}$ be a $\pi$ system and $\mathcal{L}$ be a $\lambda$-system. If $\mathcal{P} \subseteq \mathcal{L}$, then $\sigma(\mathcal{P}) \subseteq \mathcal{L}$.

Proof. Let $\lambda(\mathcal{P})$ the smallest $\lambda$-system that contains $\mathcal{P}$. We will show that $\lambda(\mathcal{P})$ is a $\sigma$-field. Then

It suffices to prove that $\lambda(\mathcal{P})$ is a $\pi$-system.

​For each $A \in \lambda (\mathcal{P})$, define

We will prove that $\Pi_A$ is a $\lambda$-system and $\Pi_A \supseteq \mathcal{P}$. It follows that $\Pi_A \supseteq \lambda(\mathcal{P})$ and $\lambda(\mathcal{P})$ is closed under intersection, and therefore a $\pi$-system.

$\Pi_A$ is a $\lambda$-system:

1. $\Omega \in \Pi_A$.

2. If $B \in \Pi_A$, we need to prove that $\bar B \in \Pi_A$. We need to prove that $A \cap \bar B = A - \big( A \cap B \big) \in \lambda(\mathcal{P})$. This follows from definition (2') of $\lambda$-system and that

   $$A, \big( A \cap B \big) \in \lambda(\mathcal{P})\, \wedge\, A \supseteq \big( A \cap B \big).$$

3. If $B_1, B_2, \ldots \in \Pi_A$ with $B_n \cap B_m = \emptyset$, then $\cup_{n \ge 1} B_n\in \Pi_A$, since

$\Pi_A$ contains $\mathcal{P}$:

If $B \in \mathcal{P}$, denote $\Pi_B = \{ A \subseteq \Omega : A \cap B \in \lambda (\mathcal{P}) \}$. Since $\mathcal{P}$ is closed under intersection, $\Pi_B \supseteq \mathcal{P}$. Since $\Pi_B$ is a $\lambda$-system, $\Pi_B \supseteq \lambda (\mathcal{P})$.

Hence, for all $A \in \lambda (\mathcal{P}), B \in \mathcal{P}$, we have $A \cap B \in \lambda (\mathcal{P})$. Therefore $\Pi_A \supseteq \mathcal{P}$.

$\square$

Unique Extension Theorem

Theorem. Let $\mathcal{P}$ be a $\pi$-system. If $\nu_1$ and $\nu_2$ are measures (on $\sigma$-fields $\mathcal{F}_1$ and $\mathcal{F}_2$) that agree on $\mathcal{P}$ and there is a sequence $A_n \in \mathcal{P}$ with $A_n \uparrow \Omega$ and $\nu_i (A_n) < \infty$ for $i = 1, 2$, then $\nu_1$ and $\nu_2$ agree on $\sigma( \mathcal{P} )$.

Proof. Since $\nu_i$ are measures on $\sigma(\mathcal{P} )$, for all $A \in \sigma(\mathcal{P})$,

It suffices to prove that $\nu_1 (A \cap A_n) = \nu_2 (A \cap A_n)$ holds for all $n \in \mathbb{N}^+$.

For a fixed $n$, now consider the set

$\triangleright$ $\mathcal{P} \subseteq \mathcal{L}:$ via assumption that $\mathcal{P}$ is a $\pi$-system and that $\nu_1$ and $\nu_2$ agrees on $\mathcal{P}$.

$\triangleright$ $\mathcal{L}$ is a $\lambda$ system:

• $\Omega \in \mathcal{L}$.

• If $B \in \mathcal{L}$, then $\bar B \in \mathcal{L}$ since

  $$\begin{aligned} \nu_1 ( \bar B \cap A_n) &= \nu_1 ( A_n ) - \nu_1 (B \cap A_n) \\ &= \nu_2 ( A_n ) - \nu_2 (B \cap A_n) = \nu_2 ( \bar B \cap A_n ), \end{aligned}$$

  where the third equality follows from that $\nu_1 ( A_n ) = \nu_2 ( A_n )$ and $\nu_1 (B \cap A_n) = \nu_2 (B \cap A_n)$.

• If $B_j$ are disjoint sets in $\mathcal{L}$, then $\cup_{j \ge 1} B_j \in \mathcal{L}$, since

  $$\begin{aligned} \nu_1 \Big( \big( \cup_{j \ge 1} B_j \big) \cap A_n \Big) &= \sum_{j} \nu_1 (B_j \cap A_n) \\ &= \sum_{j} \nu_2 (B_j \cap A_n) = \nu_2 \Big( \big( \cup_{j \ge 1} B_j \big) \cap A_n \Big). \end{aligned}$$