June 30, 2026
This is a continuation of my previous notes on measure theory and probability. Here, we will prove that, given a probability measure space \((\Omega, \mathcal{F}, P)\), there exists a unique extension of \(P\) to \(\sigma(F)\) (i.e, a probability measure on \(\sigma(F)\) ).
The theorem we will prove is: "A probability measure on a field has a unique extension to the generated \(\sigma\) -field". Specifically: Suppose \(P\) is a probability measure on a field \(F_0\) of subsets of \(\Omega\). Put \(F = \sigma(F_0)\). Then there exists a probability measure \(Q\) on \(F\) such that \(Q(A) = P(A)\) for all \(A \in F_0\). Further, if \(Q'\) is another probability measure on \(F\) such that \(Q'(A) = P(A)\) for all \(A \in F_0\), then \(Q(A) = Q'(A)\) for all \(A \in F\).
For every subset of \(A\):
\[ \begin{align*} P^*(A) = \inf \sum_n P(A_n) \end{align*} \] where the infimum is performed over all finite and infinite collections \(\{A_n\}_n\) such that \(A \subset \cup_n A_n\). We say that a set \(A\) is \(P^*\) -measurable if:
\[ \begin{align*} P^*(A \cap E) + P^*(A^c \cap E) = P^*(E) \end{align*} \]
\(P^*\) has the following properties, which we subsequently prove:
We can first show (2). \(\sum_n P(A_n) \ge 0\) since \(P(A_n) \ge 0\) for every \(A_n \in \mathcal{F}_0\) and where the \(\{A_n\}_n\) form a cover of \(A\). So, \(P^*(A) = \inf \sum_n P(A_n) \ge 0\). Next we show (1). Using (2), \(0 \le P^*(\emptyset) \le P(\emptyset) = 0\) since \(\emptyset \in \mathcal{F}_0\). So, \(P^*(\emptyset) = 0\). We now show \(3\). Since \(A \subset B\), then every cover of \(B\) is also a cover of \(A\). That is, the set of all covers of \(A\) includes the set of all covers of \(B\). Then, \(\inf \sum_n P(A_n) \le \inf \sum_n P(B_n)\), which implies that \(P^*(A) \le P^*(B)\). Let's show \(4\). We need to show that \(P^*(\cup_n A_n) \le \sum_n P^*(A_n)\) where \(A_n \cap A_m = \emptyset\) for \(n \ne m\). Let \(\epsilon > 0\) be arbitrary. Then, there exists a cover \(\{A_{n, m} \}_m\) of \(A_n\) such that:
\begin{align*} \sum_m P(A_{n,m}) &\le P^*(A_n) + \frac{\epsilon}{2^n} \\ \sum_n \sum_m P(A_{n,m}) &\le \sum_n P^*(A_n) + \epsilon & \text{ sum over $n$ } \\ P^*(\cup_n A_n) &\le \sum_n \sum_m P(A_{n,m}) \le \sum_n P^*(A_n) + \epsilon & \text{ $\cup_n A_n \subset \cup_n \cup_m A_{n,m}$ } \end{align*}We will prove a few lemmas that will be important for proving the first part of the theorem, which is that there exists an extension of the probability measure \(P\) to \(\sigma(F_0)\). We define the class \(M\) to be the collection of sets that are \(P^*\) measurable. \(A\) is \(P^*\) measurable if: \[ P^*(A \cap E) + P^*(A^c \cap E) = P^*(E) \text{ for all $E \subset \Omega$ } \]
The class \(M\) is a field.
We will need to prove 3 things:
First we prove (1): \[ P^*(\Omega \cap E) + P^*(\Omega^c \cap E) = P^*(E) + P^*(\emptyset) = P^*(E) \] Next we prove (2): Let \(A \in M\). Then we can write:
\begin{align*} P^*(E) &= P^*(A \cap E) + P^*(A^c \cap E) \\ P^*(E) &= P^*(A^c \cap E) + P^*((A^c)^c \cap E) \end{align*}The last line implies that \(A^c \in M\). Now we prove (3):
\begin{align*} P^*(E) &= P^*(A \cap E) + P^*(A^c \cap E) \\ &= P^*(A \cap B \cap E) + P^*(A \cap B^c \cap E) + P^*(A^c \cap B \cap E) + P^*(A^c \cap B^c \cap E) & \text{ $B$ is $P^*$ -measurable (see below) }\\ &\ge P^*(\{(A \cap B) \cup (A \setminus B) \cup (B \setminus A)\} \cap E) + P^*((A \cup B)^c \cap E) & \text{ countable subadditivity } \\ &= P^*((A \cup B) \cap E) + P^*((A \cup B)^c \cap E) \end{align*}The second line follows from the fact that \(B\) is \(P^*\) -measurable, so \(P^*(A \cap E) = P^*(A \cap E \cap B) + P^*(A \cap E \cap B^c)\) because \(A \cap E \subset \Omega\).
Let \(\{A_n\}_n\) be a disjoint sequence (finite or infinite) sequence of \(M\) -sets. For each \(E \subset \Omega\), \(P^*(E \cap \Bigl(\cup_n A_n\Bigl)) = \sum_n P^*(E \cap A_n)\)
The approach we take is to first prove the lemma for finitely many \(A_n\). Then, we extend it to infinite sequences. First, let \(n = 2\). Suppose \(A_1 \cup A_2 = \Omega\). So, \(A_1 = A_2^c\). So:
\begin{align*} P^*(E \cap A_1) + P^*(E \cap A_2) = P^*(E \cap A_2^c) + P^*(E \cap A_2) = P^*(E) = P^*(E \cap (A_1 \cup A_2)) \end{align*}Now, suppose \(A_1 \cup A_2 \ne \Omega\). Then, \(A_1 \in M\) and \(A_2 \in M\), so \(A_1 \cup A_2 \in M\) (by Lemma \(3.1\)). Hence:
\begin{align*} P^*(A_1 \cap E) + P^*(A_1^c \cap E) &= P^*(E) \\ P^*(A_1 \cap A_2 \cap E) + P^*(A_1 \cap A_2^c \cap E) + P^*(A_1^c \cap A_2 \cap E) + P^*(A_1^c \cap A_2^c \cap E) &= P^*(E) \\ P^*(\emptyset) + P^*(A_1 \cap E) + P^*(A_2 \cap E) + P^*(A_1^c \cap A_2^c \cap E) &= P^*(E) & A_1 \cap A_2 = \emptyset \implies A_1 \cap A_2^c = A_1, A_1^c \cap A_2 = A_2 \\ P^*(A_1 \cap E) + P^*(A_2 \cap E) &= P^*(E) - P^*(E \cap (A_1 \cup A_2)^c) \\ &\le P^*(E \cap (A_1 \cup A_2)) & \text{ $P^*$ -measurability of $A_1 \cup A_2$ } \end{align*}Since it holds for \(n = 2\), we can extend it for all \(n \ge 2\), since \(\cup_{k=1}^n A_k = \Bigl(\cup_{k=1}^{n-1} A_k\Bigl) \cup A_{n}\). Then, we can take \(n \rightarrow \infty\) to obtain the result for infinite sequences.
The class \(M\) is a \(\sigma\) -field and \(P^*\) restricted to \(M\) is countably additive
Let \(\{A_n\}_{n=1}^\infty \in M\), \(A_n \cap A_m = \emptyset\) for all \(n \ne m\). We want to show \(P^*(\cup_{n=1}^\infty) A_n = \sum_{n=1}^\infty P^*(A_n)\). By Lemma 2:
\begin{align*} P^*(\cup_{n=1}^\infty A_n) &= P^*(\Omega \cap \bigcup_{n=1}^\infty A_n) \\ &= \sum_{n=1}^\infty P^*(\Omega \cap A_n) \\ &= \sum_{n=1}^\infty P^*(A_n) \end{align*}Therefore, \(P^*\) is countably additive on \(M\). We show that if \(A_n \in M\) for all \(n \in \mathbb{N}\), then \(\cup_{n=1}^\infty A_n \in M\). We want to show that: \[P^*(E \cap \bigcup_{n=1}^\infty A_n) + P^*(E \cap \bigcap_{n=1}^\infty A_n^c) \le P^*(E)\]. First, we show that this is true for countable disjoint unions. Let \(A_m \cap A_n = \emptyset\) for all \(m \ne n\). Let \(F_n = \cup_{k=1}^n A_k\). Then:
\begin{align*} P^*(E \cap F_n) + P^*(E \cap F_n^c) &= P^*(E) & \text{$F_n \in M$ since $M$ closed under finite unions} \\ \sum_{k=1}^n P^*(E \cap A_k) + P^*(E \cap F_n^c) &\le P^*(E) & \text{Lemma 2} \end{align*}Now, \(F_n \subset \cup_{n=1}^\infty A_n\), which implies that \(\cap_{n=1}^\infty A_n^c \subset F_n^c\). So, \(P^*(E \cap \bigcap_{k=1}^\infty A_k^c) \le P^*(E \cap F_n^c)\) by monotonicity. So:
\begin{align*} \sum_{k=1}^n P^*(E \cap A_k) + P^*(E \cap \bigcap_{k=1}^\infty A_k^c) &\le P^*(E) \\ \lim_{n \rightarrow \infty} \Bigl[ \sum_{k=1}^n P^*(E \cap A_k) + P^*(E \cap \bigcap_{k=1}^\infty A_k^c) \Bigl] &\le \lim_{n \rightarrow \infty} P^*(E) \\ \sum_{k=1}^\infty P^*(E \cap A_k) + P^*(E \cap \bigcap_{k=1}^\infty A_k^c) &\le P^*(E) \\ P^*(E \cap \bigcup_{n=1}^\infty A_n) + P^*(E \cap \bigcap_{k=1}^\infty A_k^c) &\le P^*(E) \end{align*}This proves the theorem for countable disjoint unions. To show it for arbitrary unions, we can define \(B_1 = A_1\) and \(B_k = A_k \setminus \bigcup_{j=1}^{k-1} A_j\). Then, \(\cup_{k=1}^\infty B_k = \cup_{k=1}^\infty A_k\) and the \(\{B_k\}_{k=1}^\infty\) are disjoint. So, we can apply the preceding result of countable disjoint unions to the \(B_k\)'s and obtain the same result.
\(F_0 \subset M\)
Let \(A \in F_0\). We want to show that \(P^*(A \cap E) + P^*(A^c \cap E) \le P^*(E) \forall E \subset \Omega\). Let \(\epsilon > 0\). Then, there exists a cover of \(F_0\) -sets \(\{A_n\}_n\) such that \(\sum_n P(A_n) \le P^*(E) + \epsilon\). Since \(A_n \in F_0\), \(A \cap A_n \in F_0\). Also, \(E \subset \cup_n A_n\), which means \(A \cap E \subset \cup_n (A \cap A_n)\) and \(A^c \cap E \subset \cup_n (A^c \cap A_n)\). So:
\begin{align*} P^*(A \cap E) + P^*(A^c \cap E) &\le \sum_n P(A \cap A_n) + \sum_n P(A^c \cap A_n) \\ &= \sum_n P(A_n) \le P^*(E) + \epsilon & \text{ finite additivity } \end{align*}\(P^*(A) = P(A) \forall A \in F_0\)
We know from the definition of \(P^*\) that \(P^*(A) \le P(A)\). We show that \(P^*(A) \ge P(A)\). Let \(\{A_n\}_n\) be a cover of \(A\). Then, \(A_n \in F_0\) such that \(\sum_n P(A_n) \le P^*(A) + \epsilon\). Since \(A \subset \cup_n A_n\), \(A \subset \cup_n (A \cap A_n)\), and \(A \cap A_n \subset A_n\), \(\sum_n P(A \cap A_n) \le \sum_n P(A_n)\) by monotonicity. By countable subadditivity, \(P(A) = P(A \cap \cup_n A_n) \le \sum_n P(A \cap A_n) \le \sum_n P(A_n) \le P^*(A) + \epsilon\).
There exists a probability measure \(Q\) on \(F = \sigma(F_0)\) such that \(Q(A) = P(A)\) for all \(A \in F_0\).
We take \(Q = P^*\) and show it is a probability measure on \(F\) first. We know by Lemma 3 that \(M\) is a \(\sigma\) -field. By Lemma 4, \(F_0 \subset M\), so \(F = \sigma(F_0) \subset M\) by minimality of \(\sigma(F_0)\). We also know by Lemma 3 that \(Q\) restricted to \(M\) (and, therefore, to \(F\)) is countably additive. Combined with the basic properties of \(P^*\) we prove at the start of these notes (non-negativity) and the fact that \(P^*(\Omega) = P(\Omega) = 1\), this shows that \(Q\) is a probability measure on \(F\). Lemma 5 shows that \(Q(A) = P^*(A) = P(A)\) for all \(A \in F_0\), which is the second part of this theorem.
To prove uniqueness, we need some extra definitions:
A \(\pi\) -system is any class of sets \(\mathcal{P}\) where \(A, B \in \mathcal{P}\) implies \(A \cap B \in \mathcal{P}\).
A \(\lambda\) -system is any class of sets \(\mathcal{L}\) which satisfies the following criteria:
There is a small modification we can do to the definition above that will be useful for proving later results:
Given (1) and (3) hold in \(\lambda\) -system definition, condition (2) is equivalent to the following (which we'll denote (2')): If \(A, B \in \mathcal{L}\) and \(B \subset A\), \(A \setminus B \in \mathcal{L}\).
First, we show that \((2') \implies (2)\). Since \(\Omega \in \mathcal{L}\) in definition of \(\lambda\) -system, if \(A \in \mathcal{L}\), then \(A \subset \Omega\) and \(A^c = \Omega \setminus A \in \mathcal{L}\). Now, we show that \((2) \implies (2')\). Suppose \(A, B \in \mathcal{L}\) and \(B \subset A\):
\begin{align*} A \setminus B = A \cap B^c = (A^c \cup B)^c \end{align*}Note that \(B \cap A^c = \emptyset\) since \(B \cap A = A\). Then, \(A^c \cup B\) is a disjoint union, so by (3) in definition of \(\lambda\) -system, \(A^c \cup B \in \mathcal{L}\). By (2), \(A \setminus B = (A^c \cup B)^c \in \mathcal{L}\).
The next lemma relates \(\lambda\) -systems and \(\pi\) -systems to \(\sigma\) -fields:
If a class \(\mathcal{F}\) is a \(\lambda\) -system and a \(\pi\) -system, then it is a \(\sigma\) -field.
We have to show the following properties:
(1) and (2) follow from definition of \(\lambda\) -system. Also, (3) follows from definition of \(\pi\) -system. For (4), we can write \(A_1 = B_1\) and \(A_k = B_k \setminus \cup_{j=1}^{k-1} B_{j}\), and so by Turning Countable Unions to Disjoint Unions we see that the \(\{A_k\}_k\) are disjoint and \(\cup_{k=1}^\infty B_k = \cup_{k=1}^\infty A_k \in \mathcal{F}\).
If \(\mathcal{P}\) is a \(\pi\) -system and \(\mathcal{L}\) is a \(\lambda\) -system and \(\mathcal{P} \subset \mathcal{L}\), then \(\sigma(\mathcal{P}) \subset \mathcal{L}\).
The proof for this is a little intricate, so I'll lay out the strategy for showing it. Let \(M\) be the minimal \(\lambda\) -system containing \(\mathcal{P}\). Since \(\mathcal{P} \subset \mathcal{L}\) AND \(\mathcal{L}\) is a \(\lambda\) -system, \(\mathcal{M} \subset \mathcal{L}\). If we can show that \(M\) is also a \(\pi\) -system, then Lemma 8 shows that \(M\) is a \(\sigma\) -field. By minimality of \(\sigma(\mathcal{P})\), because \(\mathcal{P} \subset M \subset \mathcal{L}\), \(\sigma(\mathcal{P}) \subset M \subset \mathcal{L}\). We lay out the steps to showing that \(M\) is a \(\pi\) -system:
We now fill in the details for steps 1,2,3. First, we show that \(L_A\) is a \(\lambda\) -system. We show the 3 properties in definition of \(\lambda\) -system. First, \(A \cap \Omega \in A\), so \(\Omega \in L_A\). To show \(L_A\) is closed under complementation, we can apply Lemma 7 and prove the equivalent statement that if \(B_1, B_2 \in L_A\) and \(B_1 \subset B_2\), then \(B_2 \setminus B_1 \in L_A\). We already know that \(A \cap B_1 \in M\) and \(A \cap B_2 \in M\) and that \(A \cap B_1 \subset A \cap B_2\), so applying Lemma 7 to \(M\), \(A \cap (B_2 \setminus B_1) = (A \cap B_2) \setminus (A \cap B_1) \in M\). Let \(\{B_n\}_n\) be a collection of disjoint sets such that \(B_n \in L_A\). Then, \(A \cap B_m \in M\). So:
\begin{align*} A \cap \cup_{n=1}^\infty B_n &= \cup_{n=1}^\infty \Bigl(A \cap B_n \Bigl) \\ \end{align*}Note that \(B_n \cap B_m = \emptyset\) for all \(n \ne m\). Then, \(A \cap B_n\) and \(A \cap B_m\) are disjoint as well. So, by property 3 of definition of \(\lambda\) -system, \(\cup_{n=1}^\infty \Bigl(A \cap B_n\Bigl) \in M\).
Now, we show that step 2. If \(A, B \in \mathcal{P}\), then \(A \cap B \in \mathcal{P} \subset M\). Then, \(B \in L_A\). Hence, \(B \in \mathcal{P} \implies B \in L_A\). So, \(\mathcal{P} \subset L_A\). Since \(M\) is the minimal \(\lambda\) -system containing \(\mathcal{P}\), it follows that \(M \subset L_A\).
Step 3: If \(A \in \mathcal{P}\), then \(M \subset L_A\). That is, if \(A \in P\) and \(B \in M\), then \(B \in L_A\). This also implies \(A \in L_B\) because \(B \cap A = A \cap B \in M\). In other words, \(A \in \mathcal{P}\) and \(B \in M\) implies \(A \in L_B\). So, \(\mathcal{P} \subset L_B\). By minimality of \(M\), \(M \subset L_B\). So, if \(C \in M\), then \(C \in L_B\), which implies \(B \cap C \in M\). By definition of \(\pi\) -system, \(M\) is a \(\pi\) -system.
Now, we can prove uniqueness with the following theorem:
Let \(P_1\), \(P_2\) be 2 probability measures on \(\sigma(\mathcal{P})\) where \(\mathcal{P}\) is a \(\pi\) -system. Suppose \(P_1\) and \(P_2\) agree on \(\mathcal{P}\). Then, they agree on \(\sigma(\mathcal{P})\).
The strategy we use is to construct a \(\lambda\) -system that \(P_1\) and \(P_2\) agree on and then apply Theorem 9 to show that \(P_1\) and \(P_2\) agree on \(\sigma(\mathcal{P})\). In this case, we show that \(L = \{B \in \sigma(\mathcal{P}): P_1(B) = P_2(B)\}\) is a \(\lambda\) -system:
Let \(\{B_n\}_n\) be a collection of \(L\) -sets that are disjoint. Then, \(\cup_n B_n \in \sigma(\mathcal{P})\), so we can compute \(P_1(\cup_n B_n)\):
\begin{align*} P_1(\cup_n B_n) &= \sum_n P_1(B_n) & \text{ countable additivity } \\ &= \sum_n P_2(B_n) & \text{ $P_1(B_n) = P_2(B_n)$ } \\ &= P_2(\cup_n B_n) & \text{ countable additivity } \end{align*}So, \(\cup_n B_n \in L\).
Since every field is a \(\pi\) -system, this theorem applies for \(F_0\) and \(F = \sigma(F_0)\) since \(F_0\) is a field.
A class \(\mathcal{M}\) of subsets of \(\Omega\) is called monotone if it is closed under the formation of monotone unions and intersections:
Before we continue, there is a fact that is mentioned in the proof of the monotone class theorem (which is similar to the \(\pi\) - \(\lambda\) theorem) that we will prove here:
If \(\mathcal{F}\) is a monotone field, it is a \(\sigma\) -field.
\(\mathcal{F}\) is a monotone field. Then:
We see that (1), (2), and (3) are also properties of a \(\sigma\) -field. Now suppose that \(\{B_n\}_{n=1}^\infty\) is a sequence of sets where \(B_n \in \mathcal{F}\) for all \(n \in \mathbb{N}\). We want to show that \(\cup_{n=1}^\infty B_n \in \mathcal{F}\). Let \(A_n \in \cup_{k=1}^n B_k\) for all \(n \in \mathbb{N}\). Then, \(A_n \subset A_{n+1}\). Let \(B = \cup_{n=1}^\infty B_n\). Also, \(\cup_{n=1}^\infty A_n = \cup_{n=1}^\infty \cup_{k=1}^n B_k = \cup_{k=1}^\infty \cup_{n = k}^\infty B_k = \cup_{k=1}^\infty B_k\). So, \(B \in \mathcal{F}\) because \(A_n \uparrow A\) and \(A_n \in \mathcal{F}\) for all \(n\) by combining (2) and (3).
We have the following monotone class theorem:
If \(\mathcal{F}_0\) is a field and \(\mathcal{M}\) is a monotone class, then \(\mathcal{F}_0 \subset \mathcal{M}\) implies \(\sigma(\mathcal{F}_0) \subset \mathcal{M}\).
This proof is similar to the proof of the \(\pi\) - \(\lambda\) theorem Let \(m(\mathcal{F}_0)\) be the minimal monotone class containing \(\mathcal{F}_0\). Then, \(m(\mathcal{F}_0) \subset \mathcal{M}\). If we show that \(m(\mathcal{F}_0)\) is a field, then by Lemma 11 it is a \(\sigma\) -field. By the minimality of \(\sigma\) -fields, because \(\mathcal{F}_0 \subset m(\mathcal{F}_0) \subset \mathcal{M}\), that means that \(\sigma(\mathcal{F}_0) \subset m(\mathcal{F}_0) \subset \mathcal{M}\). So, we show now that \(m(\mathcal{F}_0)\) is a field. Like before, we will construct sets (like \(L_A\)) to obtain the requisite properties:
If \(A, B \in m(\mathcal{F}_0)\), \(A \cup B \in m(\mathcal{F}_0)\): Let:
\begin{align*} G_1 &= \{A: A \cup B \in m(\mathcal{F}_0) \forall B \in \mathcal{F}_0\} \\ G_2 &= \{B: A \cup B \in m(\mathcal{F}_0) \forall A \in m(\mathcal{F}_0) \} \end{align*}First, we show \(G_1\) and \(G_2\) are monotone.
Since \(\mathcal{F}_0 \subset G_1\), \(m(\mathcal{F}_0) \subset G_1\) by minimality (\(G_1\) is monotone). If \(B \in \mathcal{F}_0\) and \(A \in m(\mathcal{F}_0)\), then because \(m(\mathcal{F}_0) \subset G_1\), \(A \in G_1\). And, \(B \in \mathcal{F}_0 \subset m(\mathcal{F}_0) \subset G_1\). So, \(A \cup B \in m(\mathcal{F}_0)\). That is, \(B \in G_2\). Equivalently, \(\mathcal{F}_0 \subset G_2\). By minimality of \(m(\mathcal{F}_0)\) and the fact \(G_2\) is monotone, \(m(\mathcal{F}_0) \subset G_2\). In other words, if \(A, B \in m(\mathcal{F}_0)\), then \(A \cup B \in m(\mathcal{F}_0)\).
We say a probability space \((\Omega, \mathcal{F}, P)\) is complete if \(A \subset B\), \(B \in \mathcal{F}\), and \(P(B) = 0\) imply that \(A \in \mathcal{F}\).
The following statement is made in the book, but it's not immediately obvious (to me, at least):
Suppose \((\Omega, \mathcal{F}, P)\) is complete. Let \(A \in \mathcal{F}\), \(A \Delta A' \subset B \in \mathcal{F}\), and \(P(B) = 0\). Then, \(A' \in \mathcal{F}\) and \(P(A') = P(A)\).
We show first that \(A' \in \mathcal{F}\). By completeness, \(A \Delta A' \in \mathcal{F}\). By monotonicity and non-negativity of \(P\), it follows that \(P(A \Delta A') = 0\). Since \(A \in \mathcal{F}\) and \(A \Delta A' \in \mathcal{F}\), their difference, \(\Bigl(A \Delta A'\Bigl) \setminus A = A' \setminus A \in \mathcal{F}\). Using this fact, we can also deduce that \(\Bigl(A \Delta A'\Bigl) \setminus \Bigl(A' \setminus A\Bigl) = A \setminus A' \in \mathcal{F}\). Once more, we can use these facts to obtain that \(A \setminus (A \setminus A') = A \cap A' \in \mathcal{F}\). Then, we get that \(A' = (A' \cap A) \cup (A' \cap A^c)\), which is a finite union of two \(\mathcal{F}\) -sets, implying that \(A' \in \mathcal{F}\). Further, \(A = (A \cap A') \cup (A \cap (A')^c)\). Note that both \(A'\) and \(A\) are written as the union of two disjoint \(\mathcal{F}\) -sets, so by finite additivity of \(P\):
\begin{align*} P(A') &= P(A' \cap A) + P(A' \cap A^c) = P(A' \cap A) \\ P(A) &= P(A \cap A') + P(A \cap (A')^c) = P(A \cap A') = P(A' \cap A) \\ &\implies P(A) &= P(A') \end{align*}P(A' ∩ Ac) = 0 and \(P(A \cap (A')^c) = 0\) because they are two disjoint sets whose union is \((A' \cap A^c) \cup (A \cap (A')^c) = (A' \setminus A) \cup (A \setminus A') = A \Delta A'\), and \(P(A \Delta A') = 0\).
The important point about complete sets here is that for any probability space, \((\Omega, \mathcal{F}, P)\), we can enlarge the \(\sigma\) -field and extend the probability measure \(P\) to get a complete probability space, as we now show: consider the extension \(P^*\) and take \(\mathcal{F}_0 = \mathcal{F} = \sigma(\mathcal{F}_0)\). Let \(\mathcal{M}\) be the \(\sigma\) -field of \(P^*\) -measurable sets. By Lemma 3, \(P^*\) restricted to \(\mathcal{M}\) is a probability measure. We now show this space is complete. Let \(A \subset B\), \(B \in \mathcal{M}\), and \(P^*(B) = 0\). We prove that \(A \in \mathcal{M}\). Let \(E \subset \Omega\). Then:
\begin{align*} P^*(A \cap E) + P^*(A^c \cap E) \le P^*(B) + P^*(E) = P^*(E) \end{align*}where we applied monotonicity to both terms (\(A \cap E \subset B\) and \(A^c \cap E \subset E\)).
For the probability space \(((0, 1], \mathcal{B}, \lambda)\), the sets \(A\) in the completed \(\sigma\) -field \(\mathcal{M}\) are called Lebesgue sets, and \(\lambda\) is still called the Lebesgue measure.
We consider the Vitali construction, which produces a set in \((0, 1]\) that is not a Borel set (i.e, a set not in \(\mathcal{B}\)). Define, for \(x, y \in (0, 1]\): \[ x \oplus y = \begin{cases} x + y & 0 < x + y \le 1 \\ x + y - 1 & x + y > 1 \end{cases} \]
Let \(\mathcal{L}\) be the class of Borel sets \(A\) such that \(A \bigoplus x \in \mathcal{B}\) and \(\lambda(A \bigoplus x) = \lambda(A)\) for some \(x \in (0, 1]\). We show that \(\mathcal{L}\) is a \(\lambda\) -system:
Suppose \(A \in \mathcal{L}\). We show that \(A^c \in \mathcal{L}\). Since \(A \in \mathcal{B}\), \(A^c \in \mathcal{B}\) since it is a \(\sigma\) -algebra. Note that \((A \oplus x)^c = A^c \oplus x\). Since \(A \oplus x \in \mathcal{B}\), \((A \oplus x)^c \in \mathcal{B}\). Putting this together, \(A^c \oplus x \in \mathcal{B}\). Next, \(\lambda(A) = \lambda(A \oplus x)\). Since \(\Omega = (0, 1]\) and \(\lambda(A^c) = \lambda(\Omega \setminus A) = 1 - \lambda(A)\), we get that:
\begin{align*} \lambda(A^c) &= 1 - \lambda(A) \\ &= 1 - \lambda(A \oplus x) \\ &= \lambda((A \oplus x)^c) \\ &= \lambda(A^c \oplus x) \end{align*}Therefore, \(A^c \in \mathcal{L}\).
Let \(A_n \in \mathcal{L}\) and \(A_n \cap A_m = \emptyset\) for all \(m \ne n\). Then, \(A_n \in \mathcal{B}\) and \(\cup_n A_n \in \mathcal{B}\) since \(\mathcal{B}\) is a \(\sigma\) -algebra. Because, \(A \oplus x \in \mathcal{B}\), it follows that \(\cup_n (A_n \oplus x) = (\cup_n A_n) \oplus x \in \mathcal{B}\). Next:
\begin{align*} \lambda((\cup_n A_n) \oplus x) &= \lambda(\cup_n (A_n \oplus x)) \\ &= \sum_n \lambda(A_n \oplus x) & \text{ countable additivity } \\ &= \sum_n \lambda(A_n) & \text{ translation-invariance } \\ &= \sum_n \lambda(A_n) \\ &= \lambda(\cup_n A_n) & \text{ countable additivity } \end{align*}Since \(\mathcal{L}\) is a \(\lambda\) -system containing \(\mathcal{I}\), the class of intervals in \((0, 1]\), and \(\mathcal{I}\) is a field (hence a \(\pi\) -system), it follows by the \(\pi-\lambda\) theorem that \(\mathcal{B} = \sigma(I) \subset \mathcal{L}\). Define an equivalence relation \(\sim\) on \((0, 1]\) so that \(x \sim y\) if there exists \(r \in (0, 1]\) such that \(x \oplus r = y\). Let \(H\) be a subset of \((0, 1]\) consist of exactly one representative from each equivalence class (this set exists because of the axiom of choice). Since \(\mathbb{Q}\) is countable, consider the sets \(H \oplus r\) for rational \(r\). First, we show these sets are disjoint: Suppose \(H \oplus r_1\) and \(H \oplus r_2\) have a comment element \(h_r\). Then, there exists \(h_1 \in H \oplus r_1\) such that \(h_r = h_1 \oplus r_1\) and \(h_2 \in H \oplus r_2\) such that \(h_r = h_2 \oplus r_2\). Hence:
\begin{align*} h_1 \oplus r_1 &= h_2 \oplus r_2 \end{align*}If this is the case, then:
\begin{align*} h_1 + r_1 &= h_2 + r_2 & \text{ or } \\ h_1 + r_1 &= h_2 + r_2 - 1 & \text{ or } \\ h_1 + r_1 - 1 &= h_2 + r_2 & \text{ or } \\ h_1 + r_1 - 1 &= h_2 + r_2 - 1 \\ \end{align*}Either way, this would imply that \(h_1 \sim h_2\). But, \(H\) contains exactly one representative from each equivalence class, so we cannot have \(h_1 \ne h_2\), which implies that \(h_1 = h_2\). But this would imply that \(r_1 = r_2\). Hence, the sets are actually the same. Therefore, two distinct sets \(H \oplus r_1\) and \(H \oplus r_2\) are disjoint. Therefore, we can write \((0, 1] = \cup_{r \in \mathbb{Q}} H \oplus r\) because for every \(y \in (0, 1]\), then there exists some \(r \in \mathbb{Q}\) so that \(h \oplus r = y\) for some \(h \in H\) since \(H\) contains one member from each equivalence class. Now, suppose that \(H\) were in \(\mathcal{B}\). Then, \(\lambda((0, 1]) = \sum_r \lambda(H \oplus r)\) since the \(H \oplus r\) are disjoint and the union is countable because \(\mathcal{Q}\) is countable. However, we know that \(H \in \mathcal{B}\) implies \(H \in \mathcal{L}\), which means that \(\lambda(H) = \lambda(H \oplus r)\). So, the sum \(\sum_r \lambda(H) = \lambda((0, 1]) = 1\). Since \(\lambda(H) \ge 0\) and the sum is over a countably infinite quantity, either \(\lambda(H) = 0\) (which means \(1 = 0\)) or \(\lambda(H) > 0\) in which case we have a divergent sum (and so it cannot equal \(1\)). This means \(H\) is not a Borel set.
Part A: We use finite additivity explicitly in Lemma 4 and implicitly in Lemma 5 since we invoke monotonicity of \(P\), which uses finite additivity in its proof. We also invoke countable subadditivity of \(P\) in Lemma 5, which is a property that is derived from the countable additivity of \(P\).
Part B: Consider the set \(\mathbb{N}\) and define a set function \(P\) so that for any subset \(A\) of \(\mathbb{N}\), \(P(A) = 0\) if \(A\) is finite and \(1\) if \(A\) is infinite. The collection of all subsets of \(\mathbb{N}\) forms a \(\sigma\) -field, so \(P\) is a probability measure. \(P\) is finitely additive since for any finite set \(A\) of \(\mathbb{N}\), \(P(A) = 0\) for every \(x \in A\), and \(\sum_{x \in A} P(\{x\}) = \sum_{x \in A} 0 = 0 = P(A)\). Write \(\mathbb{N} = \cup_{n \in \mathbb{N}} \{n\}\). We don't have countable subadditivity because:
\begin{align*} 1 = P(\mathbb{N}) = P(\cup_{n \in \mathbb{N}} \{n\}) \not\le \sum_{n \in \mathbb{N}} P(\{n\}) = 0. \end{align*}Now, suppose that \((\Omega, \mathcal{F}, P)\) is a probability space and suppose \(P\) is a probability measure that is finitely additive and countably subadditive. Let $\{An\}n=1^∞ be a collection of disjoint \(\mathcal{F}\) -sets. Let \(A = \cup_{n=1}^\infty A_n \in \mathcal{F}\). We prove that \(P(A) = P(\cup_{n=1}^\infty A_n) = \sum_{n=1}^\infty P(A_n)\) (it is countably additive). By countable subadditivity, \(P(A) \le \sum_{n=1}^\infty P(A_n)\). We show that \(P(A) \ge \sum_{n=1}^\infty P(A_n)\). Let \(n \in \mathbb{N}\) be fixed. Then, for every \(A_k \in \mathcal{F}\), \(k \in [n]\), \(\cup_{k=1}^\infty A_k \in \mathcal{F}\) since \(\mathcal{F}\) is a field. Also, \(\cup_{k=1}^n A_k \subset A\). Since \(A \in \mathcal{F}\) and \(\cup_{k=1}^n A_k \in \mathcal{F}\), it follows that \(A \setminus \cup_{k=1}^n A_k \in \mathcal{F}\). By finite additivity of \(P\), \(P(\cup_{k=1}^n A_k) + P(A \setminus \cup_{k=1}^n A_k) = P(A)\). Since \(P(A \setminus \cup_{k=1}^n A_k) \ge 0\), it follows that \(P(A) - P(\cup_{k=1}^n A_k) \ge 0\). Again, by finite additivity, \(P(\cup_{k=1}^n A_k) = \sum_{k=1}^n P(A_k)\), so \(P(A) - \sum_{k=1}^n P(A_k) \ge 0\). Then, \(\lim_{n \rightarrow \infty} \Bigl[P(A) - \sum_{k=1}^n P(A_k)\Bigl] \ge 0\), which implies \(P(A) \ge \sum_{k=1}^\infty P(A_k)\). Therefore, \(P(A) = \sum_{k=1}^\infty P(A_k)\).
Part C: Lemma 5 uses countable subadditivity for \(P\) on a field \(\mathcal{F}_0\), so if Theorem 2.1 were weakened to only apply to \(\sigma\) -fields, we could not apply it to \(\mathcal{F}_0\).
Part A: We know that there exists a unique extension of the probability measure \(P\) on \(\mathcal{F_0}\) to \(\mathcal{F}\) and that this extension is \(P^*\) itself, so we will prove that:
\begin{align} P^*(A) &= \inf [P^*(B): A \subset B, B \in \mathcal{F}] \\ P_*(B) &= \sup [P^*(C): C \subset A, C \in \mathcal{F}] \end{align}and prove that the supremum and infimum are achieved. First, we show that (1) is true. Let \(S = \{P^*(B): A \subset B, B \in \mathcal{F}\}\). Let \(l = \inf S\). By monotonicity of \(P^*\), \(P^*(A) \le P^*(B)\) for all \(B \in \mathcal{F}\) with \(A \subset B\), so \(P^*(A) \le l\). We now show that \(P^*(A) \ge l\). Let \(\epsilon > 0\). There exists a collection of \(\mathcal{F}\) -sets, \(\{A_n\}_n\), satisfying \(A \subset \cup_n A_n\) so that \(\sum_n P(A_n) \le P^*(A) + \epsilon\). By Lemma 5, \(P(A_n) = P^*(A_n)\), so \(\sum_n P^*(A_n) \le P^*(A) + \epsilon\). By countable subadditivity of \(P^*\), \(P^*(A_n) \le P^*(A) + \epsilon\). Since \(\mathcal{F}_0 \subset \mathcal{F}\), each \(A_n \in \mathcal{F}\) and \(\cup_n A_n \subset \mathcal{F}\). So, \(P^*(\cup_n A_n) \in S\) and \(l \le P^*(\cup_n A_n)\). Therefore, \(l \le P^*(A) + \epsilon\). Since \(\epsilon\) was arbitrary, it follows that \(l \le P^*(A)\). For every \(k \in \mathbb{N}\) and arbitrary \(\epsilon > 0\), there exists \(A_k\) where \(A \subset A_k\) and \(A_k \in \mathcal{F}\) so that \(P^*(A_k) \le P^*(A) + \frac{\epsilon}{2^k}\). Since \(A \subset \cap_k A_k \in \mathcal{F}\), \(P^*(\cap_k A_k) \in S\). So, \(P^*(A) \le P^*(\cap_k A_k)\). And, $P*(∩n An) ≤ \(P^*(A_k)\) for every \(k \in \mathbb{N}\), so \(P^*(\cap_n A_n) \le \sum_k P^*(A_k) \le P^*(A) + \sum_k \frac{\epsilon}{2^k} = P^*(A) + \epsilon\). Since \(\epsilon > 0\) was arbitrary, \(P^*(\cap_n A_n) = P^*(A)\) and \(\cap_n A_n \in S\). Therefore, the infimum is always achieved. We now show the analogous results for the supremum, \(\sup [P^*(C): C \subset A, C \in \mathcal{F}]\). Let \(C \in \mathcal{F}\) with \(C \subset A\) be an arbitrary set. Then, put \(B = C^c\). Since \(\mathcal{F}\) is a field, \(B \in \mathcal{F}\). Also, \(A^c \subset B\). Also, \(P^*\) is a probability measure on \(\mathcal{F}\), so \(P^*(C) = 1 - P^*(B)\). Since \(C\) was arbitrary:
\begin{align*} \sup [P^*(C): C \subset A, C \in \mathcal{F}] &= \sup [1 - P^*(B): A^c \subset B, B \in \mathcal{F}] \\ &= 1 + \sup [-P^*(B): A^c \subset B, B \in \mathcal{F}] \\ &= 1 - \inf [P^*(B): A^c \subset B, B \in \mathcal{F}] \\ &= 1 - P^*(A^c) \\ &= P_*(A) \end{align*}Part B: Let \(A\) be \(P^*\) -measurable. Then, for every \(E \subset \Omega\), \(P^*(A \cap E) + P^*(A^c \cap E) = P^*(E)\). If we let \(E = \Omega\), \(P^*(A) + P^*(A^c) = 1\), which implies \(P^*(A) = P_*(A)\). Now suppose \(P^*(A) = P_*(A) = 1 - P^*(A^c)\). Let \(E \subset \Omega\). We only need to show \(P^*(A \cap E) + P^*(A^c \cap E) \le P^*(E)\). By part (a), \(P^*(A) = P^*(B)\) where \(A \subset B\) and \(B \in \mathcal{F}\) and \(P_*(A) = P^*(C)\) where \(C \subset A\) and \(C \in \mathcal{F}\). Since \(B \in \mathcal{F}\) and \(C \in \mathcal{F}\), both \(B\) and \(C\) are \(P^*\) -measurable. So:
\begin{align*} P^*(A \cap E) + P^*(A^c \cap E) &\le P^*(B \cap E) + P^*(C^c \cap E) \\ &= P^*(B \cap E) + P^*(B \cap C^c \cap E) + P^*(B^c \cap C^c \cap E) \\ &\le P^*(B \cap E) + P^*(B \cap C^c \cap E) + P^*(B^c \cap E) \\ &= P^*(E) + P^*(B \cap C^c \cap E) \\ &= P^*(E) + P^*((B \setminus C) \cap E) \end{align*}Since \(P^*\) is countably (and hence finitely) additive, \(P^*(B) = P^*(B \setminus C) + P^*(C)\). Since \(P^*(B) = P^*(C)\), \(P^*(B \setminus C) = 0\). So, \(P^*((B \setminus C) \cap E) \le P^*(B \setminus C) = 0\) by monotonicity of \(P^*\). By non-negativity of \(P^*\), \(P^*((B \setminus C) \cap E) = 0\) by monotonicity. So, \(P^*(A \cap E) + P^*(A^c \cap E) \le P^*(E)\).
Based on the preceding lemmas, \(P\) will only agree with \(P^*\) if and only if \(P\) is countably additive. The probability measures given in (a) through (e) are all from problems 2.13 and problems 2.15, where we determined whether the probability measures were countably additive or not. Only for parts (b) and (e) do \(P\) and \(P^*\) not agree. For part (f): Let \(\{A_n\}_n\) be a countable collection of \(\mathcal{F}_0\) -sets that are disjoint. Consider \(A = \cup_n A_n\). If \(P(A) = 1\), \(\omega_0 \in \cup_n A_n\). Then, because they are a disjoint collection, there is exactly one \(A_k\), \(k \in [n]\), so that \(\omega_0 \in A_k\). So, \(P(A_k) = 1\) and \(P(A_n) = 0\) for all \(n \ne k\). If \(P(A) = 0\), then \(\omega_0 \not\in A_n\) for all \(n\), so \(P(A_n) = 0\). We know that there exists \(A \in \mathcal{F_0}\) for which \(P(A) = 1\) in the first place because \(\{\omega_0\} \in \sigma(\mathcal{F}_0)\), implying that \(\{\omega_0\} \in \mathcal{F}_0\) because if no set in \(\mathcal{F}_0\) contains \(\omega_0\), then there exists a \(\sigma\) -field not containing \(\omega_0\) that contains \(\mathcal{F}_0\), which would imply \(\{\omega_0\} \not\in \sigma(\mathcal{F}_0)\), a contradiction. Therefore, \(P(\cup_n A_n) = \sum_n P(A_n)\), and \(P\) agrees with \(P^*\).
We first show that \(P^*(A)\) is an outer measure. \(P^*(\emptyset) = f(\lambda^*(\emptyset)) = f(0) = 0\). \(f\) is non-negative since it is defined on \([0, \infty)\), \(f(0) = 0\), and \(f\) is strictly increasing, meaning \(f(x) \ge 0\) for all \(x\) in its domain. So, \(P^*(A) \ge 0\). Let \(A \subset B\). Then, by monotonicity of \(\lambda^*\), \(\lambda^*(A) \le \lambda^*(B)\). If \(\lambda^*(A) < \lambda^*(B)\), then \(f(\lambda^*(A)) < f(\lambda^*(B))\), which implies \(P^*(A) < P^*(B)\). If \(\lambda^*(A) = \lambda^*(B)\), then \(f(\lambda^*(A)) = f(\lambda^*(B))\), so \(P^*(A) = P^*(B)\). Therefore, \(P^*(A) \le P^*(B)\). Let \(\{A_n\}_n\) be a collection of \(\mathcal{B}\) -sets. We show that \(P^*(\cup_n A_n) \le \sum_n P^*(A_n)\). By countable additivity of \(\lambda^*\), \(\lambda^*(\cup_n A_n) \le \sum_n \lambda^*(A_n)\). Then, \(f(\lambda^*(\cup_n A_n)) \le f(\sum_n \lambda^*(A_n))\). We must show \(f(\sum_n \lambda^*(A_n)) < \sum_n f(\lambda^*(A_n))\). Consider the case for \(n = 2\). Then, we must show that:
\begin{align*} f(\lambda^*(A_1) + \lambda^*(A_2)) < f(\lambda^*(A_1)) + f(\lambda^*(A_2)) \end{align*}We have:
\begin{align*} f(\lambda^*(A_1)) + f(\lambda^*(A_2)) &= f\Bigl(\frac{\lambda^*(A_1)}{\lambda^*(A_1) + \lambda^*(A_2)} \cdot (\lambda^*(A_1) + \lambda^*(A_2))\Bigl) + f\Bigl(\frac{\lambda^*(A_2)}{\lambda^*(A_1) + \lambda^*(A_2)} \cdot (\lambda^*(A_1) + \lambda^*(A_2))\Bigl) \\ &= f\Bigl(\frac{\lambda^*(A_1)}{\lambda^*(A_1) + \lambda^*(A_2)} \cdot (\lambda^*(A_1) + \lambda^*(A_2)) + \frac{\lambda^*(A_2)}{\lambda^*(A_1) + \lambda^*(A_2)} \cdot 0\Bigl) + f\Bigl(\frac{\lambda^*(A_2)}{\lambda^*(A_1) + \lambda^*(A_2)} \cdot (\lambda^*(A_1) + \lambda^*(A_2)) + \frac{\lambda^*(A_1)}{\lambda^*(A_1) + \lambda^*(A_2)} \cdot 0\Bigl) &> f\Bigl(\lambda^*(A_1) + \lambda^*(A_2)\Bigl) \end{align*}by concavity of \(f\) and the fact that \(f(0) = 0\). By induction on \(n\), we can extend the result to get \(f(\lambda^*(\cup_n A_n)) < \sum_n f(\lambda^*(A_n))\). This implies that \(P^*(\cup_n A_n) \le \sum_n P^*(A_n)\). Let \(A\) be \(P^*\) -measurable. Then, \(P^*(A) + P^*(A^c) = f(1)\). That is, \(f(\lambda^*(A)) + f(\lambda^*(A^c)) = f(1)\). Since \(f\) is strictly concave, if \(\lambda^*(A) > 0\) and \(\lambda^*(A^c) > 0\), then \(f(\lambda^*(A) + \lambda^*(A^c)) < f(\lambda^*(A)) + f(\lambda^*(A^c)) = f(1)\). This implies \(f(1) < f(1)\), a contradiction. So \(\lambda^*(A) = 0\) or \(\lambda^*(A^c) = 0\). Now suppose \(\lambda^*(A) = 0\) or \(\lambda^*(A^c) = 0\). Let \(E \subset \Omega\). Suppose WLOG \(\lambda^*(A) = 0\). Then, we must show that \(P^*(A \cap E) + P^*(A^c \cap E) \le P^*(E)\). We have that \(P^*(A \cap E) = f(\lambda^*(A \cap E)) \le f(\lambda^*(A)) = 0\). Then, \(f(\lambda^*(A \cap E)) = 0\) by non-negativity, so \(\lambda^*(A \cap E) = 0\). This implies that \(P^*(A \cap E) = 0\). Then, since \(A^c \cap E \subset E\) and \(P^*\) is monotonic, \(P^*(A^c \cap E) \le P^*(E)\). Suppose this \(P^*\) does arise from definition (3.1). Then, \(P^*\) should be countably additive. Let \(P\) and \(\mathcal{F}_0\) be a probability measure and field, respectively, so that \(P^*(A) = \inf \sum_n P(A_n) = f(\lambda^*(A))\) (clearly, \(\mathcal{F}_0\) must consist of subsets of \((0, 1]\) for \(f(\lambda^*(A))\) to be defined). Let \(\{A_n\}_n\) be a collection of \(\mathcal{F}\) -sets (\(\mathcal{F} = \sigma(\mathcal{F}_0)\)) that are disjoint. Then, \(P^*(\cup_n A_n) = f(\lambda^*(\cup_n A_n)) < \sum_n f(\lambda^*(A_n))\) since \(f\) is strictly concave and because \(\lambda^*(\cup_n A_n) = \sum_n \lambda^*(A_n)\) and \(f(\sum_n \lambda^*(A_n)) < \sum_n f(\lambda^*(A_n))\). So it cannot be that \(P^*\) arises from (3.1), because it is not countably additive.
The \(\mathcal{F}\) -set \([(x, y): x \in (0, 1], 0 < y \le 1] = [(x, y): 0 < x \le 1, 0 < y \le 1]\) is the unit square \(\Omega\), so \(\Omega \in \mathcal{F}\). Suppose \(C \in \mathcal{F}\). Then, \(C = [(x, y): x \in A, 0 < y \le 1]\) for some \(A \in \mathcal{B}\). Then, since \(\mathcal{B}\) is a field, \(A^c \in \mathcal{B}\), and:
\begin{align*} C^c = [(x, y): x \not\in A, 0 < y \le 1] = [(x, y): x \in A^c, 0 < y \le 1] \\ \end{align*}Therefore, \(C^c \in \mathcal{F}\). Let \(A, B \in \mathcal{F}\). Then:
\begin{align*} A &= [(x, y): x \in A', 0 < y \le 1] \\ B &= [(x, y): x \in B', 0 < y \le 1] \end{align*}for some \(A', B' \in \mathcal{B}\). Since \(\mathcal{B}\) is a field, \(A' \cap B' \in \mathcal{B}\). So, the set \([(x, y): x \in A' \cap B', 0 < y \le 1] = A \cap B\) is in \(\mathcal{F}\). Therefore, \(\mathcal{F}\) is a field. We show \(P\) is a probability measure on \(\mathcal{F}\). For every set \(A \in \mathcal{F}\), \(A = [(x, y): x \in A', 0 < y \le 1]\) for some \(A' \in \mathcal{B}\) and \(P(A) = \lambda(A')\). So, \(P(A) \ge 0\) and \(P(\Omega) = 1\). Let \(\{A_n\}_n\) be a collection of disjoint \(\mathcal{F}\) -sets. Then:
\begin{align*} A_n = [(x, y): x \in A_n', 0 < y \le 1] \end{align*}for some \(A_n' \in \mathcal{B}\). Since \(\mathcal{B}\) is a \(\sigma\) -field, \(\cup_n A_n' \in \mathcal{B}\), so the set:
\begin{align*} \cup_n A_n = [(x, y): x \in \cup_n A_n', 0 < y \le 1] \end{align*}is in \(\mathcal{F}\). So, \(P(\cup_n A_n) = \lambda(\cup_n A_n') = \sum_n \lambda(A_n') = \sum_n P(A_n)\). Let \(A = [(x, y): 0 < x \le 1, y = \frac{1}{2}]\). Then, \(A \subset [(x, y): 0 < x \le 1, 0 < y \le 1] = \Omega\). Then, if \(\{B_n\}_n\) is a sequence of \(\mathcal{F}\) -sets so that \(A \subset \cup_n B_n\), then each \(B_n = [(x, y): x \in B_n', 0 < y \le 1]\) for some \(B_n' \in \mathcal{B}\), so we would need \((0, 1] \subset \cup_n B_n'\). But \(\cup_n B_n' \subset (0, 1]\), so \(\cup_n B_n' = (0, 1]\). This implies that \(\cup_n B_n = \Omega\). Since the \(\{B_n\}_n\) were an arbitrary cover, it follows that \(P^*(A) = 1\). Then, \(A^c = [(x, y): x \in (0, 1], y \in (0, 1] \setminus \frac{1}{2}]\). Let \(\{B_n\}_n\) be a collection of \(\mathcal{F}\) -sets so that \(A^c \subset \cup_n B_n\). Again, each \(B_n = [(x, y): x \in B_n', 0 < y \le 1]\) for some \(B_n' \in \mathcal{B}\), so \(\cup_n B_n' = (0, 1]\), so \(\cup_n B_n = \Omega\). Therefore, \(P^*(A^c) = 1\), implying that \(P_*(A^c) = 0\).
We can write the Cantor set as \(\cap_{n=1}^\infty C_n\) where \(C_1 = (0, \frac{1}{3}) \cup (\frac{2}{3}, 1)\), \(C_2 = (0, \frac{1}{9}) \cup (\frac{2}{9}, \frac{1}{3}) \cup (\frac{2}{3}, \frac{7}{9}) \cup (\frac{8}{9}, 1)\), etc. That is: \(C_n\) is defined to be the union of the disjoint intervals remaining after we remove the middle thirds of each of the intervals making up \(C_{n-1}\). For every \(n \in \mathbb{N}\), \(C_n\) is composed of \(2^n\) disjoint intervals each of length \(\frac{1}{3^n}\), so countable additivity yields that \(\lambda(C_n) = (\frac{2}{3})^n\). By monotonicity, \(\lambda(\cap_{n=1}^\infty C_n) \le \lambda(C_k) = (\frac{2}{3})^k\) for all \(k \in \mathbb{N}\). This can only be satisfied if \(\lambda(\cap_{n=1}^\infty C_n) = 0\).