149 L04 : Extending Probability Measures to Countable Additivity

Video 149.1: Lesson 4: Extending a probability measure from a field to the generated \sigma-field.

149.1 Lesson map

Recap: measure, probability measure, and finite additivity.
Goal: extend a probability measure from known sets to more sets.
Start with a field \mathcal{F}_0 and the generated \sigma-field \mathcal{F}=\sigma(\mathcal{F}_0).
Infimum and supremum: minimum/maximum ideas when endpoints may not be included.
Infimum as greatest lower bound; supremum as least upper bound.
The \varepsilon trick for infimum and supremum arguments.
Cover arbitrary subsets of \Omega using sets from \mathcal{F}_0.
Define the associated outer measure P^*.
Basic properties of P^*: empty set, monotonicity, and agreement direction on \mathcal{F}_0.
Countable subadditivity of P^*.
Open questions: is P^* a measure, a probability measure, and does it preserve P on \mathcal{F}_0?

149.2 Goal of the lesson

The previous lesson defined measures and probability measures on a measurable space. This lesson begins the harder construction problem:

Suppose a probability measure is already defined on a smaller collection of sets. Can we extend it to a larger \sigma-field without changing the values we already assigned?

The running setup is:

\Omega \neq \emptyset, \qquad \mathcal{F}_0 \text{ is a field on } \Omega, \qquad \mathcal{F}=\sigma(\mathcal{F}_0).

A field of sets is closed under complements and finite unions. A \sigma-field is closed under complements and countable unions. So \mathcal{F}_0 is almost a \sigma-field, but may fail countable-union closure.

The slide lists three assumptions: Omega is a non-empty set, F naught is a field on Omega, and F is the sigma-field generated by F naught. — Starting with a field and its generated sigma-field.

149.3 Probability measure on a field

A probability measure on a \sigma-field satisfies:

P(\emptyset)=0;
if A_1,A_2,\ldots are disjoint, then

P\left(\bigcup_{n=1}^{\infty}A_n\right) = \sum_{n=1}^{\infty}P(A_n);

P(\Omega)=1.

The subtlety is that \mathcal{F}_0 is only a field. If A_1,A_2,\ldots\in\mathcal{F}_0, the countable union \bigcup_n A_n may not belong to \mathcal{F}_0. So countable additivity is only meaningful when the union is still in the domain.

The slide writes P of A equals a number sign, then lists three rules: P of the empty set is zero, countable additivity for disjoint sets, and P of Omega equals one. — Probability-measure axioms written for sets in the field.

The goal is to extend P from \mathcal{F}_0 to the larger \sigma-field \mathcal{F}, ideally without changing the original values of P.

The slide says that P will be extended to a probability measure on all of F. A diagram shows smaller known sets inside a larger space, with additional sets suggested in the generated sigma-field. — Extending P to a probability measure on all of the generated sigma-field.

149.4 Infimum and supremum

The outer-measure construction uses infimums. The idea is close to a minimum, except the best lower bound need not itself be inside the set.

For example,

S=(0,1).

This set has no minimum and no maximum, but it has

\inf S=0, \qquad \sup S=1.

The slide introduces infimums and supremums using the set of real numbers strictly between zero and one, then asks what its minimum and maximum are. — The open interval from zero to one has no minimum or maximum.

The slide states that the infimum of the set of real numbers strictly between zero and one is zero, and the supremum is one. — Infimum and supremum of the open unit interval.

Definition 149.1 (Infimum and supremum) Let S\subseteq\mathbb{R}.

The infimum of S, written \inf S, is the greatest lower bound of S.

The supremum of S, written \sup S, is the least upper bound of S.

The slide labels the infimum as the greatest lower bound and the supremum as the least upper bound, illustrated on the open interval from zero to one. — Infimum as greatest lower bound and supremum as least upper bound.

If the minimum or maximum exists, then the infimum or supremum agrees with it.

The slide states that the infimum of the closed unit interval equals its minimum, zero, and the supremum equals its maximum, one. — When the endpoints are included, infimum equals minimum and supremum equals maximum.

149.5 The epsilon trick

A recurring proof move in analysis is the \varepsilon characterization of infimum and supremum.

If m=\inf S, then for every \varepsilon>0, there is some s\in S such that

s < m+\varepsilon.

Otherwise, m+\varepsilon would still be a lower bound, contradicting the claim that m was the greatest lower bound.

The slide illustrates that for any positive epsilon, there is a point s in the set S below the infimum plus epsilon. — The epsilon property of the infimum.

Similarly, if M=\sup S, then for every \varepsilon>0, there is some s\in S such that

s > M-\varepsilon.

The slide illustrates that for any positive epsilon, there is a point s in the set S above the supremum minus epsilon. — The epsilon property of the supremum.

This \varepsilon idea is the technical lever in the proof of countable subadditivity for P^*.

149.6 Covering arbitrary subsets of \Omega

Now return to the extension problem. Let

A\subseteq\Omega.

The set A need not belong to \mathcal{F}_0 or even to \mathcal{F}=\sigma(\mathcal{F}_0). To estimate its size, cover it using sets from \mathcal{F}_0, whose probabilities are already known.

The slide shows Omega as a large region, a subset A inside it, and asks what P of A should be. It assumes F naught is a field, F is generated by F naught, and P is a probability measure on F naught. — An arbitrary subset A of Omega whose probability is not yet known.

A cover means

A\subseteq \bigcup_{n=1}^{\infty} A_n, \qquad A_n\in\mathcal{F}_0.

Such covers always exist because \Omega\in\mathcal{F}_0 and A\subseteq\Omega.

The slide shows the set A covered by several overlapping sets from F naught inside Omega. — Covering A by sets from the field.

149.7 The associated outer measure

Definition 149.2 (Outer measure associated with P) For any A\subseteq\Omega, define

P^*(A) = \inf \left\{ \sum_{n=1}^{\infty}P(A_n) : A_n\in\mathcal{F}_0 \text{ and } A\subseteq\bigcup_{n=1}^{\infty}A_n \right\}.

In words: cover A by countably many sets from \mathcal{F}_0, add the probabilities of those covering sets, and take the smallest possible upper value in the infimum sense.

The slide defines the outer measure associated with P as P star of A equals the infimum of the sum of P of A n, taken over all sequences in F naught that cover A. — Definition of the outer measure associated with P.

More compactly,

P^*(A) = \inf\left\{ \sum_{n=1}^{\infty}P(A_n) : A_n\in\mathcal{F}_0,\; A\subseteq\bigcup_{n=1}^{\infty}A_n \right\}.

At this stage, P^* is only a candidate extension tool. The lesson proves that it has several measure-like properties, but postpones the full extension result.

The slide restates the formula for P star and begins listing properties. The first property is P star of the empty set equals zero, proved by taking all covering sets to be empty. — First property of P star: the empty set has outer measure zero.

149.8 Basic properties of P^*

Proposition 149.1 (Empty set) P^*(\emptyset)=0.

Proof: cover \emptyset by A_n=\emptyset for all n. Then

\sum_{n=1}^{\infty}P(A_n)=0.

Since all candidate sums are non-negative, the infimum is 0.

Proposition 149.2 (Monotonicity) If A\subseteq B\subseteq\Omega, then

P^*(A)\leq P^*(B).

Any cover of B is also a cover of A. Since A has at least as many admissible covers as B, the infimum for A cannot be larger than the infimum for B.

The slide states that if A is a subset of B, then P star of A is less than or equal to P star of B. The proof says any covering of B is also a covering of A. — Monotonicity of the outer measure.

Proposition 149.3 (Upper bound on original field values) If A\in\mathcal{F}_0, then

P^*(A)\leq P(A).

Proof: cover A by the sequence

A_1=A, \qquad A_2=A_3=\cdots=\emptyset.

Then

P^*(A) \leq P(A)+P(\emptyset)+P(\emptyset)+\cdots = P(A).

149.9 Countable subadditivity of P^*

The main technical result in this lesson is that P^* is countably subadditive.

Proposition 149.4 (Countable subadditivity) For any sequence of subsets A_1,A_2,\ldots\subseteq\Omega,

P^*\left(\bigcup_{n=1}^{\infty}A_n\right) \leq \sum_{n=1}^{\infty}P^*(A_n).

The slide states countable subadditivity for P star: for any sets A1, A2 and so on inside Omega, P star of their union is at most the sum of the P star values. — Countable subadditivity of P star.

149.9.1 Proof idea

For each A_n, choose a countable cover by sets from \mathcal{F}_0:

A_n \subseteq \bigcup_{k=1}^{\infty} A_{nk}.

The double index means: n selects the set being covered, and k selects a covering set for that particular A_n.

The slide begins the proof of countable subadditivity. It notes that each A n can be covered with F naught sets A n1, A n2 and so on. A note says at least one sequence exists by taking each covering set to be Omega. — Each A n can be covered by a sequence of F naught sets.

Now use the infimum \varepsilon trick. For any \varepsilon>0, choose the cover of A_n so that its total probability is within \varepsilon2^{-n} of the infimum:

\sum_{k=1}^{\infty}P(A^*_{nk}) < P^*(A_n)+\varepsilon 2^{-n}.

The slide chooses, for each A n, a covering sequence from F naught such that the sum of probabilities is less than P star of A n plus epsilon times two to the negative n. — Using the epsilon trick to choose nearly optimal covers.

Summing over n gives

\begin{aligned} \sum_{n=1}^{\infty}\sum_{k=1}^{\infty}P(A^*_{nk}) & < \sum_{n=1}^{\infty} \left(P^*(A_n)+\varepsilon 2^{-n}\right) \\ & = \sum_{n=1}^{\infty}P^*(A_n)+\varepsilon. \end{aligned}

The slide sums the nearly optimal cover inequalities over n and warns that infinite sums must be handled carefully, especially when divergent positive and negative terms are involved. — A warning about distributing infinite sums.

The slide simplifies the double sum bound by pulling out epsilon and using the geometric series sum of two to the negative n, resulting in the sum of P star of A n plus epsilon. — The geometric error terms sum to epsilon.

On the other hand,

\bigcup_{n=1}^{\infty}A_n \subseteq \bigcup_{n=1}^{\infty}\bigcup_{k=1}^{\infty}A^*_{nk}.

So the double-indexed family is a cover of the whole union. By definition of P^*,

P^*\left(\bigcup_{n=1}^{\infty}A_n\right) \leq \sum_{n=1}^{\infty}\sum_{k=1}^{\infty}P(A^*_{nk}).

The slide shows that if each A n is covered by the union over k of A n k star, then the union over n of A n is covered by the double union over n and k. It then applies the definition of the infimum to bound P star of the union. — The double-indexed cover covers the union of the A n.

Combining the two inequalities,

P^*\left(\bigcup_{n=1}^{\infty}A_n\right) < \sum_{n=1}^{\infty}P^*(A_n)+\varepsilon.

Because \varepsilon>0 was arbitrary,

P^*\left(\bigcup_{n=1}^{\infty}A_n\right) \leq \sum_{n=1}^{\infty}P^*(A_n).

The slide completes the proof: P star of the countable union is less than the double sum, which is less than the sum of P star of A n plus epsilon. Letting epsilon go to zero gives the desired inequality. — Letting epsilon go to zero completes countable subadditivity.

149.10 What has been proved so far?

Starting with a probability measure P on a field \mathcal{F}_0, we defined an outer measure candidate on every subset of \Omega:

P^*(A) = \inf \left\{ \sum_{n=1}^{\infty}P(A_n): A_n\in\mathcal{F}_0,\; A\subseteq\bigcup_{n=1}^{\infty}A_n \right\}.

We proved:

P^*(\emptyset)=0;
if A\subseteq B, then P^*(A)\leq P^*(B);
if A\in\mathcal{F}_0, then P^*(A)\leq P(A);
P^* is countably subadditive.

149.11 What remains open?

The lesson ends with the natural extension questions:

Is P^* actually a measure?
Is it a probability measure?
Does it preserve the original values on the field, meaning

P^*(A)=P(A) \qquad \text{for all } A\in\mathcal{F}_0?

Only one direction, P^*(A)\leq P(A), was shown in this lesson. The rest is deferred to the next lesson.

149.12 Takeaway

The strategy is to extend by covering from above:

\text{known probabilities on } \mathcal{F}_0 \quad\longrightarrow\quad \text{covers of arbitrary } A\subseteq\Omega \quad\longrightarrow\quad P^*(A)=\inf \text{ cover-sums}.

The infimum chooses the best upper estimate among all countable covers. This turns the extension problem into a covering problem, which is the key move behind the later construction of a probability measure on \sigma(\mathcal{F}_0).

--- title: "L04 : Extending Probability Measures to Countable Additivity" subtitle: "Measure Theoretic Probability - Jem Corcoran" date: 2026-04-17 keywords: [measure theory, probability measures, sigma algebra, outer measure, infimum, supremum] --- ::: {#vid-C99-L04 .column-margin} {{< video https://www.youtube.com/watch?v=BGMGR6q06rY >}} Lesson 4: Extending a probability measure from a field to the generated $\sigma$-field. ::: ## Lesson map - Recap: measure, probability measure, and finite additivity. - Goal: extend a probability measure from known sets to more sets. - Start with a field $\mathcal{F}_0$ and the generated $\sigma$-field $\mathcal{F}=\sigma(\mathcal{F}_0)$. - *Infimum* and *supremum*: minimum/maximum ideas when endpoints may not be included. - *Infimum* as greatest lower bound; *supremum* as least upper bound. - The $\varepsilon$ trick for *infimum* and *supremum* arguments. - Cover arbitrary subsets of $\Omega$ using sets from $\mathcal{F}_0$. - Define the associated outer measure $P^*$. - Basic properties of $P^*$: empty set, monotonicity, and agreement direction on $\mathcal{F}_0$. - Countable subadditivity of $P^*$. - Open questions: is $P^*$ a measure, a probability measure, and does it preserve $P$ on $\mathcal{F}_0$? ## Goal of the lesson The previous lesson defined measures and probability measures on a measurable space. This lesson begins the harder construction problem: > Suppose a probability measure is already defined on a smaller collection of sets. Can we extend it to a larger $\sigma$-field without changing the values we already assigned? The running setup is: $$ \Omega \neq \emptyset, \qquad \mathcal{F}_0 \text{ is a field on } \Omega, \qquad \mathcal{F}=\sigma(\mathcal{F}_0). $$ A **field** of sets is closed under complements and finite unions. A **$\sigma$-field** is closed under complements and countable unions. So $\mathcal{F}_0$ is almost a $\sigma$-field, but may fail countable-union closure. ::: {.column-margin #fig-L04-slide-01} ![Starting with a field and its generated sigma-field.](images/l04-s01.png){fig-alt="The slide lists three assumptions: Omega is a non-empty set, F naught is a field on Omega, and F is the sigma-field generated by F naught." group="slides"} The construction starts from a field $\mathcal{F}_0$ on $\Omega$ and defines the larger measurable collection $\mathcal{F}=\sigma(\mathcal{F}_0)$. ::: ## Probability measure on a field A probability measure on a $\sigma$-field satisfies: 1. $P(\emptyset)=0$; 2. if $A_1,A_2,\ldots$ are disjoint, then $$ P\left(\bigcup_{n=1}^{\infty}A_n\right) = \sum_{n=1}^{\infty}P(A_n); $$ 3. $P(\Omega)=1$. The subtlety is that $\mathcal{F}_0$ is only a field. If $A_1,A_2,\ldots\in\mathcal{F}_0$, the countable union $\bigcup_n A_n$ may not belong to $\mathcal{F}_0$. So countable additivity is only meaningful when the union is still in the domain. ::: {.column-margin #fig-L04-slide-02} ![Probability-measure axioms written for sets in the field.](images/l04-s02.png){fig-alt="The slide writes P of A equals a number sign, then lists three rules: P of the empty set is zero, countable additivity for disjoint sets, and P of Omega equals one." group="slides"} When $P$ is defined only on $\mathcal{F}_0$, the right-hand side $\sum_n P(A_n)$ always makes sense for $A_n\in\mathcal{F}_0$, but the left-hand side $P(\bigcup_n A_n)$ only makes sense if the countable union is also in $\mathcal{F}_0$. ::: The goal is to extend $P$ from $\mathcal{F}_0$ to the larger $\sigma$-field $\mathcal{F}$, ideally without changing the original values of $P$. ::: {.column-margin #fig-L04-slide-03} ![Extending P to a probability measure on all of the generated sigma-field.](images/l04-s03.png){fig-alt="The slide says that P will be extended to a probability measure on all of F. A diagram shows smaller known sets inside a larger space, with additional sets suggested in the generated sigma-field." group="slides"} The intended extension should keep the old probabilities on $\mathcal{F}_0$ while assigning probabilities to the new sets in $\mathcal{F}=\sigma(\mathcal{F}_0)$. ::: ## Infimum and supremum The outer-measure construction uses **infimums**. The idea is close to a minimum, except the best lower bound need not itself be inside the set. For example, $$ S=(0,1). $$ This set has no minimum and no maximum, but it has $$ \inf S=0, \qquad \sup S=1. $$ ::: {.column-margin #fig-L04-slide-04} ![The open interval from zero to one has no minimum or maximum.](images/l04-s04.png){fig-alt="The slide introduces infimums and supremums using the set of real numbers strictly between zero and one, then asks what its minimum and maximum are." group="slides"} For the open interval $(0,1)$, neither endpoint belongs to the set. So there is no minimum or maximum, even though $0$ and $1$ are the natural boundary values. ::: ::: {.column-margin #fig-L04-slide-05} ![Infimum and supremum of the open unit interval.](images/l04-s05.png){fig-alt="The slide states that the infimum of the set of real numbers strictly between zero and one is zero, and the supremum is one." group="slides"} The infimum and supremum recover the boundary values: $\inf(0,1)=0$ and $\sup(0,1)=1$, even though $0,1\notin(0,1)$. ::: ::: {#def-infimum-supremum} ## Infimum and supremum Let $S\subseteq\mathbb{R}$. The **infimum** of $S$, written $\inf S$, is the greatest lower bound of $S$. The **supremum** of $S$, written $\sup S$, is the least upper bound of $S$. ::: ::: {.column-margin #fig-L04-slide-06} ![Infimum as greatest lower bound and supremum as least upper bound.](images/l04-s06.png){fig-alt="The slide labels the infimum as the greatest lower bound and the supremum as the least upper bound, illustrated on the open interval from zero to one." group="slides"} For $(0,1)$, every number $\leq 0$ is a lower bound, but $0$ is the greatest one. Every number $\geq 1$ is an upper bound, but $1$ is the least one. ::: If the minimum or maximum exists, then the infimum or supremum agrees with it. ::: {.column-margin #fig-L04-slide-07} ![When the endpoints are included, infimum equals minimum and supremum equals maximum.](images/l04-s07.png){fig-alt="The slide states that the infimum of the closed unit interval equals its minimum, zero, and the supremum equals its maximum, one." group="slides"} For the closed interval $[0,1]$, the endpoints belong to the set, so $\inf[0,1]=\min[0,1]=0$ and $\sup[0,1]=\max[0,1]=1$. ::: ## The epsilon trick A recurring proof move in analysis is the $\varepsilon$ characterization of infimum and supremum. If $m=\inf S$, then for every $\varepsilon>0$, there is some $s\in S$ such that $$ s < m+\varepsilon. $$ Otherwise, $m+\varepsilon$ would still be a lower bound, contradicting the claim that $m$ was the greatest lower bound. ::: {.column-margin #fig-L04-slide-08} ![The epsilon property of the infimum.](images/l04-s08.png){fig-alt="The slide illustrates that for any positive epsilon, there is a point s in the set S below the infimum plus epsilon." group="slides"} The infimum can be approached from inside the set: once we move slightly above $\inf S$, we must encounter a point of $S$. ::: Similarly, if $M=\sup S$, then for every $\varepsilon>0$, there is some $s\in S$ such that $$ s > M-\varepsilon. $$ ::: {.column-margin #fig-L04-slide-09} ![The epsilon property of the supremum.](images/l04-s09.png){fig-alt="The slide illustrates that for any positive epsilon, there is a point s in the set S above the supremum minus epsilon." group="slides"} The supremum can be approached from inside the set: once we move slightly below $\sup S$, we must encounter a point of $S$. ::: This $\varepsilon$ idea is the technical lever in the proof of countable subadditivity for $P^*$. ## Covering arbitrary subsets of $\Omega$ Now return to the extension problem. Let $$ A\subseteq\Omega. $$ The set $A$ need not belong to $\mathcal{F}_0$ or even to $\mathcal{F}=\sigma(\mathcal{F}_0)$. To estimate its size, cover it using sets from $\mathcal{F}_0$, whose probabilities are already known. ::: {.column-margin #fig-L04-slide-10} ![An arbitrary subset A of Omega whose probability is not yet known.](images/l04-s10.png){fig-alt="The slide shows Omega as a large region, a subset A inside it, and asks what P of A should be. It assumes F naught is a field, F is generated by F naught, and P is a probability measure on F naught." group="slides"} For $A\subseteq\Omega$, the value $P(A)$ may not be defined. The construction therefore compares $A$ with countable covers by sets $A_n\in\mathcal{F}_0$. ::: A cover means $$ A\subseteq \bigcup_{n=1}^{\infty} A_n, \qquad A_n\in\mathcal{F}_0. $$ Such covers always exist because $\Omega\in\mathcal{F}_0$ and $A\subseteq\Omega$. ::: {.column-margin #fig-L04-slide-11} ![Covering A by sets from the field.](images/l04-s11.png){fig-alt="The slide shows the set A covered by several overlapping sets from F naught inside Omega." group="slides"} The covering sets may overlap. The construction intentionally sums their probabilities anyway, producing an upper estimate for the unknown size of $A$. ::: ## The associated outer measure ::: {#def-associated-outer-measure} ## Outer measure associated with $P$ For any $A\subseteq\Omega$, define $$ P^*(A) = \inf \left\{ \sum_{n=1}^{\infty}P(A_n) : A_n\in\mathcal{F}_0 \text{ and } A\subseteq\bigcup_{n=1}^{\infty}A_n \right\}. $$ ::: In words: cover $A$ by countably many sets from $\mathcal{F}_0$, add the probabilities of those covering sets, and take the smallest possible upper value in the infimum sense. ::: {.column-margin #fig-L04-slide-12} ![Definition of the outer measure associated with P.](images/l04-s12.png){fig-alt="The slide defines the outer measure associated with P as P star of A equals the infimum of the sum of P of A n, taken over all sequences in F naught that cover A." group="slides"} The outer measure $P^*(A)$ is defined for every subset $A\subseteq\Omega$, not just for sets in $\mathcal{F}_0$ or $\mathcal{F}$. ::: More compactly, $$ P^*(A) = \inf\left\{ \sum_{n=1}^{\infty}P(A_n) : A_n\in\mathcal{F}_0,\; A\subseteq\bigcup_{n=1}^{\infty}A_n \right\}. $$ At this stage, $P^*$ is only a candidate extension tool. The lesson proves that it has several measure-like properties, but postpones the full extension result. ::: {.column-margin #fig-L04-slide-13} ![First property of P star: the empty set has outer measure zero.](images/l04-s13.png){fig-alt="The slide restates the formula for P star and begins listing properties. The first property is P star of the empty set equals zero, proved by taking all covering sets to be empty." group="slides"} The empty set can be covered by $\emptyset,\emptyset,\ldots$, and each has probability zero. Therefore $P^*(\emptyset)=0$. ::: ## Basic properties of $P^*$ ::: {#prp-pstar-empty} ## Empty set $$ P^*(\emptyset)=0. $$ ::: Proof: cover $\emptyset$ by $A_n=\emptyset$ for all $n$. Then $$ \sum_{n=1}^{\infty}P(A_n)=0. $$ Since all candidate sums are non-negative, the infimum is $0$. ::: {#prp-pstar-monotone} ## Monotonicity If $A\subseteq B\subseteq\Omega$, then $$ P^*(A)\leq P^*(B). $$ ::: Any cover of $B$ is also a cover of $A$. Since $A$ has at least as many admissible covers as $B$, the infimum for $A$ cannot be larger than the infimum for $B$. ::: {.column-margin #fig-L04-slide-14} ![Monotonicity of the outer measure.](images/l04-s14.png){fig-alt="The slide states that if A is a subset of B, then P star of A is less than or equal to P star of B. The proof says any covering of B is also a covering of A." group="slides"} Monotonicity follows directly from the cover definition: covering the larger set $B$ automatically covers the smaller set $A$. ::: ::: {#prp-pstar-upper-bound-on-field} ## Upper bound on original field values If $A\in\mathcal{F}_0$, then $$ P^*(A)\leq P(A). $$ ::: Proof: cover $A$ by the sequence $$ A_1=A, \qquad A_2=A_3=\cdots=\emptyset. $$ Then $$ P^*(A) \leq P(A)+P(\emptyset)+P(\emptyset)+\cdots = P(A). $$ ::: {.column-margin #fig-L04-slide-15} ![For sets already in the field, P star is at most P.](images/l04-s15.png){fig-alt="The slide states that for any A in F naught, P star of A is less than or equal to P of A. It covers A with A followed by empty sets, then sums the probabilities." group="slides"} For original sets $A\in\mathcal{F}_0$, one admissible cover is $A,\emptyset,\emptyset,\ldots$. This proves $P^*(A)\leq P(A)$, but equality still requires more work. ::: ## Countable subadditivity of $P^*$ The main technical result in this lesson is that $P^*$ is countably subadditive. ::: {#prp-pstar-countably-subadditive} ## Countable subadditivity For any sequence of subsets $A_1,A_2,\ldots\subseteq\Omega$, $$ P^*\left(\bigcup_{n=1}^{\infty}A_n\right) \leq \sum_{n=1}^{\infty}P^*(A_n). $$ ::: ::: {.column-margin #fig-L04-slide-16} ![Countable subadditivity of P star.](images/l04-s16.png){fig-alt="The slide states countable subadditivity for P star: for any sets A1, A2 and so on inside Omega, P star of their union is at most the sum of the P star values." group="slides"} This property resembles countable additivity, but it does not require the $A_n$ to be disjoint and gives an inequality rather than equality. ::: ### Proof idea For each $A_n$, choose a countable cover by sets from $\mathcal{F}_0$: $$ A_n \subseteq \bigcup_{k=1}^{\infty} A_{nk}. $$ The double index means: $n$ selects the set being covered, and $k$ selects a covering set for that particular $A_n$. ::: {.column-margin #fig-L04-slide-17} ![Each A n can be covered by a sequence of F naught sets.](images/l04-s17.png){fig-alt="The slide begins the proof of countable subadditivity. It notes that each A n can be covered with F naught sets A n1, A n2 and so on. A note says at least one sequence exists by taking each covering set to be Omega." group="slides"} Each $A_n$ has at least one admissible cover because we can always take $A_{nk}=\Omega$ for all $k$. ::: Now use the infimum $\varepsilon$ trick. For any $\varepsilon>0$, choose the cover of $A_n$ so that its total probability is within $\varepsilon2^{-n}$ of the infimum: $$ \sum_{k=1}^{\infty}P(A^*_{nk}) < P^*(A_n)+\varepsilon 2^{-n}. $$ ::: {.column-margin #fig-L04-slide-18} ![Using the epsilon trick to choose nearly optimal covers.](images/l04-s18.png){fig-alt="The slide chooses, for each A n, a covering sequence from F naught such that the sum of probabilities is less than P star of A n plus epsilon times two to the negative n." group="slides"} The factor $2^{-n}$ splits the total error budget across the sequence. Since $\sum_{n=1}^{\infty}2^{-n}=1$, the accumulated error will be at most $\varepsilon$. ::: Summing over $n$ gives $$ \begin{aligned} \sum_{n=1}^{\infty}\sum_{k=1}^{\infty}P(A^*_{nk}) & < \sum_{n=1}^{\infty} \left(P^*(A_n)+\varepsilon 2^{-n}\right) \\ & = \sum_{n=1}^{\infty}P^*(A_n)+\varepsilon. \end{aligned} $$ ::: {.column-margin #fig-L04-slide-19} ![A warning about distributing infinite sums.](images/l04-s19.png){fig-alt="The slide sums the nearly optimal cover inequalities over n and warns that infinite sums must be handled carefully, especially when divergent positive and negative terms are involved." group="slides"} The proof can distribute the sum here because all terms are non-negative. This avoids invalid manipulations such as treating divergent positive and negative infinite sums as if they were ordinary finite numbers. ::: ::: {.column-margin #fig-L04-slide-20} ![The geometric error terms sum to epsilon.](images/l04-s20.png){fig-alt="The slide simplifies the double sum bound by pulling out epsilon and using the geometric series sum of two to the negative n, resulting in the sum of P star of A n plus epsilon." group="slides"} The error terms collapse neatly: $\varepsilon\sum_{n=1}^{\infty}2^{-n}=\varepsilon$. ::: On the other hand, $$ \bigcup_{n=1}^{\infty}A_n \subseteq \bigcup_{n=1}^{\infty}\bigcup_{k=1}^{\infty}A^*_{nk}. $$ So the double-indexed family is a cover of the whole union. By definition of $P^*$, $$ P^*\left(\bigcup_{n=1}^{\infty}A_n\right) \leq \sum_{n=1}^{\infty}\sum_{k=1}^{\infty}P(A^*_{nk}). $$ ::: {.column-margin #fig-L04-slide-21} ![The double-indexed cover covers the union of the A n.](images/l04-s21.png){fig-alt="The slide shows that if each A n is covered by the union over k of A n k star, then the union over n of A n is covered by the double union over n and k. It then applies the definition of the infimum to bound P star of the union." group="slides"} The double union is just one large countable cover of $\bigcup_n A_n$, so the outer measure of the union is no greater than the total probability of this cover. ::: Combining the two inequalities, $$ P^*\left(\bigcup_{n=1}^{\infty}A_n\right) < \sum_{n=1}^{\infty}P^*(A_n)+\varepsilon. $$ Because $\varepsilon>0$ was arbitrary, $$ P^*\left(\bigcup_{n=1}^{\infty}A_n\right) \leq \sum_{n=1}^{\infty}P^*(A_n). $$ ::: {.column-margin #fig-L04-slide-22} ![Letting epsilon go to zero completes countable subadditivity.](images/l04-s22.png){fig-alt="The slide completes the proof: P star of the countable union is less than the double sum, which is less than the sum of P star of A n plus epsilon. Letting epsilon go to zero gives the desired inequality." group="slides"} The proof ends by letting the arbitrary error $\varepsilon$ shrink to zero. This gives countable subadditivity of $P^*$. ::: ## What has been proved so far? Starting with a probability measure $P$ on a field $\mathcal{F}_0$, we defined an outer measure candidate on every subset of $\Omega$: $$ P^*(A) = \inf \left\{ \sum_{n=1}^{\infty}P(A_n): A_n\in\mathcal{F}_0,\; A\subseteq\bigcup_{n=1}^{\infty}A_n \right\}. $$ We proved: 1. $P^*(\emptyset)=0$; 2. if $A\subseteq B$, then $P^*(A)\leq P^*(B)$; 3. if $A\in\mathcal{F}_0$, then $P^*(A)\leq P(A)$; 4. $P^*$ is countably subadditive. ## What remains open? The lesson ends with the natural extension questions: - Is $P^*$ actually a measure? - Is it a probability measure? - Does it preserve the original values on the field, meaning $$ P^*(A)=P(A) \qquad \text{for all } A\in\mathcal{F}_0? $$ Only one direction, $P^*(A)\leq P(A)$, was shown in this lesson. The rest is deferred to the next lesson. ## Takeaway The strategy is to extend by **covering from above**: $$ \text{known probabilities on } \mathcal{F}_0 \quad\longrightarrow\quad \text{covers of arbitrary } A\subseteq\Omega \quad\longrightarrow\quad P^*(A)=\inf \text{ cover-sums}. $$ The infimum chooses the best upper estimate among all countable covers. This turns the extension problem into a covering problem, which is the key move behind the later construction of a probability measure on $\sigma(\mathcal{F}_0)$.