153 L08 : Limits of Sequences of Sets

Video 153.1: Lesson 8: limits of sequences of sets.

153.1 Lesson map

This lesson introduces limits of sequences of sets.

0:12 — What should \lim_n A_n mean for sets?
2:03 — First intuitive example: increasing intervals approaching [0,1).
3:24 — Review: \limsup and \liminf for sequences of real numbers.
10:29 — Decreasing sequences of sets: A_n\downarrow A.
12:06 — Increasing sequences of sets: A_n\uparrow A.
13:39 — General sequences: define \limsup A_n and \liminf A_n.
17:09 — Prove \liminf A_n\subseteq\limsup A_n.
19:35 — Define the limit of sets when \liminf A_n=\limsup A_n.
20:00 — Check decreasing sequences.
23:39 — Check increasing sequences.
24:18 — Revisit the interval example.
27:22 — Alternating-set example.
30:09 — Complements of \limsup and \liminf.
31:46 — Event interpretation: infinitely often and eventually.
34:09 — Measurability of \limsup A_n, \liminf A_n, and \lim A_n.
35:36 — Preview: continuity of probabilities.

153.2 Why limits of sets?

So far, collections like A_1,A_2,A_3,\ldots were mostly just indexed families. The order did not matter much when taking unions or intersections.

From now we begin to care about the order. How does the sequence of sets evolve over time - we might be dealing with a time series, their index being time. Or they may evolve so that an events may depend on a previous one. In any case, we want to understand the long-term behavior of a sequence of sets. A sequence of sets is meant to evolve: A_1, A_2, A_3, \ldots The question is whether this sequence has a limiting set.

The slide introduces the question of defining the limit of a sequence of sets A n. It shows a sequence of circular sets moving or shrinking toward a possible limiting object. — A sequence of sets may appear to approach a limiting set.

153.3 First example: nested intervals

Let \Omega=\mathbb{R},\quad A_n=\left[0,1-\frac{1}{n+1}\right]. Then

A_1=\left[0,\frac12\right], \qquad A_2=\left[0,\frac23\right], \qquad A_3=\left[0,\frac34\right], \quad \ldots

The sets are increasing: A_1\subseteq A_2\subseteq A_3\subseteq\cdots., intuitively, they approach [0,1)

Increasing closed intervals approaching the half-open interval [0,1)

This example is geometrically clear, but we need a definition that also works for arbitrary measurable sets.

153.4 Warm-up: \limsup and \liminf of real sequences

Before defining limits of sets, recall the analogous ideas for a real sequence a_1,a_2,a_3,\ldots. The limit superior or limit supremum tracks the eventual upper envelope of the sequence.

The slide shows a plotted real sequence with points A1, A2, A3 and so on, visually clustered between an upper and lower curve. — A real sequence oscillating between upper and lower envelopes.

For each n, look at the tail \{a_m:m\geq n\}. Take the supremum of that tail: \sup_{m\geq n} a_m. As n increases, we discard more initial terms, so these tail suprema form a non-increasing sequence. The limit of these tail suprema is the \limsup:

\limsup_{n\to\infty} a_n = \lim_{n\to\infty} \sup_{m\geq n} a_m \tag{153.1}

Equivalently,

\limsup_{n\to\infty} a_n = \inf_{n\geq 1}\sup_{m\geq n}a_m. \tag{153.2}

The slide explains limsup for a sequence of real numbers by removing initial terms and recomputing the supremum of the remaining tail. The resulting tail suprema form a non-increasing sequence. — The limsup is the limit of tail suprema.

Similarly, the limit inferior or limit infimum tracks the eventual lower envelope:

\liminf_{n\to\infty} a_n = \lim_{n\to\infty} \inf_{m\geq n}a_m = \sup_{n\geq 1}\inf_{m\geq n}a_m.

The slide explains liminf by removing initial terms and recomputing the infimum of the remaining tail. The resulting tail infima form a non-decreasing sequence. — The liminf is the limit of tail infima.

If the two envelopes meet, the sequence has a limit:

\limsup_{n\to\infty} a_n = \liminf_{n\to\infty} a_n = L.

Then

\lim_{n\to\infty} a_n=L.

The slide states that if the limsup and liminf of a real sequence are equal, their common value is the limit of the sequence. — A real sequence has a limit when limsup and liminf agree.

The set version will imitate this idea, but with unions and intersections replacing suprema and infima.

153.5 Monotone sequences of sets

153.5.1 Decreasing sets

Suppose

A_1\supseteq A_2\supseteq A_3\supseteq\cdots.

Then the sets are decreasing. The natural limiting set is the part that remains forever:

A=\bigcap_{n=1}^{\infty}A_n.

We write

A_n\downarrow A.

The slide shows nested sets A1 containing A2 containing A3 and so on. It suggests that the limiting set is the intersection of all A n. — A decreasing sequence of sets converges to the intersection.

153.5.2 Increasing sets

Suppose

A_1\subseteq A_2\subseteq A_3\subseteq\cdots.

Then the sets are increasing. The natural limiting set is everything that eventually appears:

A=\bigcup_{n=1}^{\infty}A_n.

We write

A_n\uparrow A.

The slide shows nested sets A1 inside A2 inside A3 and so on. It suggests that the limiting set is the union of all A n. — An increasing sequence of sets converges to the union.

These two cases motivate the general definitions.

153.6 General sequences of sets

Now let

A_1,A_2,A_3,\ldots

be any sequence of sets. They need not be increasing, decreasing, disjoint, or nested.

To define the set-theoretic \limsup and \liminf, use tails of the sequence.

153.6.1 Tail unions

For each n, define the tail union

\bigcup_{m=n}^{\infty}A_m.

As n increases, we remove more initial sets, so the tail unions decrease:

\bigcup_{m=1}^{\infty}A_m \supseteq \bigcup_{m=2}^{\infty}A_m \supseteq \bigcup_{m=3}^{\infty}A_m \supseteq \cdots.

The limit of this decreasing sequence is its intersection.

The slide forms the tail unions of a general sequence of sets: the union of all A m from m equals 1 onward, then from m equals 2 onward, then from m equals 3 onward. These tail unions form a decreasing sequence. — Tail unions form a decreasing sequence of sets.

153.6.2 Tail intersections

For each n, define the tail intersection

\bigcap_{m=n}^{\infty}A_m.

As n increases, we remove more initial constraints, so the tail intersections increase:

\bigcap_{m=1}^{\infty}A_m \subseteq \bigcap_{m=2}^{\infty}A_m \subseteq \bigcap_{m=3}^{\infty}A_m \subseteq \cdots.

The limit of this increasing sequence is its union.

The slide forms the tail intersections of a general sequence of sets: the intersection of all A m from m equals 1 onward, then from m equals 2 onward, then from m equals 3 onward. These tail intersections form an increasing sequence. — Tail intersections form an increasing sequence of sets.

153.7 Definitions: \limsup and \liminf of sets

Definition 153.1 (Limit superior of sets) For a sequence of sets A_1,A_2,\ldots, define

\limsup_{n\to\infty} A_n = \bigcap_{n=1}^{\infty} \bigcup_{m=n}^{\infty}A_m.

Definition 153.2 (Limit inferior of sets) For a sequence of sets A_1,A_2,\ldots, define

\liminf_{n\to\infty} A_n = \bigcup_{n=1}^{\infty} \bigcap_{m=n}^{\infty}A_m.

The slide defines limsup A n as the intersection over n of the union over m at least n of A m, and liminf A n as the union over n of the intersection over m at least n of A m. — Definitions of limsup and liminf for sequences of sets.

Both are sets. They are made entirely from countable unions and intersections.

153.8 Basic containment

For every sequence of sets,

\liminf_{n\to\infty}A_n \subseteq \limsup_{n\to\infty}A_n.

The slide states that the liminf of a sequence of sets is contained in the limsup of the sequence, and begins a proof by taking an element omega in the liminf. — The liminf set is contained in the limsup set.

Proof sketch:

Take

\omega\in\liminf A_n = \bigcup_{n=1}^{\infty}\bigcap_{m=n}^{\infty}A_m.

Then for some fixed N,

\omega\in\bigcap_{m=N}^{\infty}A_m.

So \omega\in A_m for every m\geq N. Therefore, for every n, \omega belongs to the tail union

\bigcup_{m=n}^{\infty}A_m.

Hence

\omega\in \bigcap_{n=1}^{\infty} \bigcup_{m=n}^{\infty}A_m = \limsup A_n.

153.9 Definition of the limit of sets

Definition 153.3 (Limit of a sequence of sets) If

\liminf_{n\to\infty}A_n = \limsup_{n\to\infty}A_n,

then the common set is called the limit of the sequence, and we write

\lim_{n\to\infty}A_n = \liminf_{n\to\infty}A_n = \limsup_{n\to\infty}A_n.

If the two sets are not equal, then the sequence of sets does not have a limit.

The slide defines the limit of a sequence of sets as the common value when the liminf and limsup are equal. — A sequence of sets has a limit when liminf and limsup agree.

153.10 Checking the monotone cases

153.10.1 Decreasing case

Suppose

A_1\supseteq A_2\supseteq A_3\supseteq\cdots.

Then

\limsup A_n = \bigcap_{n=1}^{\infty}A_n,

because each tail union is simply its first set:

\bigcup_{m=n}^{\infty}A_m=A_n.

The slide computes limsup for a decreasing sequence of sets and shows that each tail union equals A n, so the limsup becomes the intersection of all A n. — For decreasing sets, limsup is the intersection.

Similarly,

\liminf A_n = \bigcap_{n=1}^{\infty}A_n.

Thus,

A_n\downarrow A \quad\Rightarrow\quad \lim_n A_n=A=\bigcap_{n=1}^{\infty}A_n.

The slide computes liminf for a decreasing sequence and shows it also equals the intersection of all A n. Therefore the limit exists and equals the intersection. — For decreasing sets, liminf is also the intersection.

153.10.2 Increasing case

A_1\subseteq A_2\subseteq A_3\subseteq\cdots,

then the analogous result is

A_n\uparrow A \quad\Rightarrow\quad \lim_n A_n=A=\bigcup_{n=1}^{\infty}A_n.

The slide asks the viewer to verify the increasing case: if A n increases, then limsup and liminf both equal the union of all A n. — For increasing sets, the limit is the union.

153.11 Revisit the interval example

Recall

A_n=\left[0,1-\frac{1}{n+1}\right].

This is an increasing sequence, so the limit is the union:

\lim_{n\to\infty}A_n = \bigcup_{n=1}^{\infty} \left[0,1-\frac{1}{n+1}\right] = [0,1).

![The interval example has limit 0,1).

Figure 153.18: The endpoint 1 is approached but never included by any A_n, so it is not in the limiting set.

153.12 Alternating sets

Consider a sequence alternating between two sets:

A_n= \begin{cases} B, & n \text{ odd},\\ C, & n \text{ even}. \end{cases}

Then every tail union contains both B and C, so

\limsup A_n = B\cup C.

Every tail intersection contains only what is common to both, so

\liminf A_n = B\cap C.

Therefore the limit exists only if

B\cup C = B\cap C,

which happens when B=C.

The slide gives a coin-flipping-inspired example where A n alternates between two sets B and C. It computes limsup as B union C and liminf as B intersect C. — An alternating sequence has limsup B union C and liminf B intersect C.

153.13 Complements

The complement of a \limsup is a \liminf of complements:

\left(\limsup_{n\to\infty}A_n\right)^c = \liminf_{n\to\infty}A_n^c.

Using the definition,

\left( \bigcap_{n=1}^{\infty} \bigcup_{m=n}^{\infty}A_m \right)^c = \bigcup_{n=1}^{\infty} \bigcap_{m=n}^{\infty}A_m^c.

This is just De Morgan’s law applied to countable intersections and unions.

Likewise,

\left(\liminf_{n\to\infty}A_n\right)^c = \limsup_{n\to\infty}A_n^c.

The slide applies De Morgan's laws to show that the complement of limsup A n equals liminf of A n complement, and vice versa. — Complements swap limsup and liminf.

153.14 Event language

The \limsup has a useful event interpretation.

A point \omega is in

\limsup A_n = \bigcap_{n=1}^{\infty}\bigcup_{m=n}^{\infty}A_m

exactly when, no matter how far out in the sequence we go, \omega appears in some later A_m.

So \omega is in infinitely many of the A_n.

This is often written

\limsup A_n = [A_n\ \text{i.o.}],

where i.o. means infinitely often.

The slide explains that omega in limsup A n means omega belongs to infinitely many of the A n. It introduces the notation A n i.o., meaning infinitely often. — The limsup event means A n occurs infinitely often.

The \liminf means eventual membership:

\liminf A_n = \bigcup_{n=1}^{\infty}\bigcap_{m=n}^{\infty}A_m.

A point \omega is in \liminf A_n when there is some finite time N after which

\omega\in A_m \qquad \text{for all } m\geq N.

That is, \omega belongs to all but finitely many of the A_n.

The slide explains that omega in liminf A n means omega is in all but finitely many of the A n. It mentions the notation A n eventually. — The liminf event means A n occurs eventually.

153.15 Measurability of set limits

Suppose (\Omega,\mathcal{F},P) is a probability space and

A_n\in\mathcal{F} \qquad \text{for all } n.

Because \mathcal{F} is a \sigma-field, it is closed under countable unions and intersections. Therefore,

\limsup A_n = \bigcap_{n=1}^{\infty} \bigcup_{m=n}^{\infty}A_m \in\mathcal{F},

and

\liminf A_n = \bigcup_{n=1}^{\infty} \bigcap_{m=n}^{\infty}A_m \in\mathcal{F}.

So it makes sense to discuss

P(\limsup A_n), \qquad P(\liminf A_n).

The slide states that if A n are in a sigma-field F, then limsup A n and liminf A n are also in F because they are built using countable unions and intersections. — Limsup and liminf of measurable sets are measurable.

If the limit exists, then

\lim_n A_n\in\mathcal{F},

P\left(\lim_{n\to\infty}A_n\right)

is also well-defined.

The slide says that if liminf and limsup agree, then the limit set belongs to the sigma-field, so the probability of the limit set is defined. — The probability of a set limit is well-defined when the limit exists.

153.16 Preview: continuity of probabilities

The final question is whether probability commutes with limits of sets.

Compare

P\left(\lim_{n\to\infty}A_n\right)

with

\lim_{n\to\infty}P(A_n).

These are different-looking operations:

first take a set limit, then measure it;
first measure each set, then take a numerical limit.

The next lesson studies when these agree. This is called continuity of probabilities.

The slide asks how P of the limit of A n relates to the limit of P of A n, previewing continuity of probabilities and the Borel-Cantelli lemmas. — Preview of continuity of probabilities.

153.17 Takeaway

The key definitions are:

\limsup_{n\to\infty}A_n = \bigcap_{n=1}^{\infty}\bigcup_{m=n}^{\infty}A_m,

\liminf_{n\to\infty}A_n = \bigcup_{n=1}^{\infty}\bigcap_{m=n}^{\infty}A_m.

The interpretation is:

\limsup A_n = \{\omega:\omega\in A_n \text{ infinitely often}\},

\liminf A_n = \{\omega:\omega\in A_n \text{ eventually}\}.

The limit exists when these two sets are equal.

--- title: "L08 : Limits of Sequences of Sets" subtitle: "Measure Theoretic Probability - Jem Corcoran" date: 2026-04-17 keywords: [measure theory, probability, sigma algebra, limits of sets, limsup, liminf] --- ::: {#vid-C99-l08 .column-margin} {{< video https://youtu.be/7mo9tAm4oow?list=PLLyj1Zd4UWrO6VtBSiQLsNlo9QBm30nxC >}} Lesson 8: limits of sequences of sets. ::: ## Lesson map This lesson introduces limits of sequences of sets. - 0:12 — What should $\lim_n A_n$ mean for sets? - 2:03 — First intuitive example: increasing intervals approaching $[0,1)$. - 3:24 — Review: $\limsup$ and $\liminf$ for sequences of real numbers. - 10:29 — Decreasing sequences of sets: $A_n\downarrow A$. - 12:06 — Increasing sequences of sets: $A_n\uparrow A$. - 13:39 — General sequences: define $\limsup A_n$ and $\liminf A_n$. - 17:09 — Prove $\liminf A_n\subseteq\limsup A_n$. - 19:35 — Define the limit of sets when $\liminf A_n=\limsup A_n$. - 20:00 — Check decreasing sequences. - 23:39 — Check increasing sequences. - 24:18 — Revisit the interval example. - 27:22 — Alternating-set example. - 30:09 — Complements of $\limsup$ and $\liminf$. - 31:46 — Event interpretation: infinitely often and eventually. - 34:09 — Measurability of $\limsup A_n$, $\liminf A_n$, and $\lim A_n$. - 35:36 — Preview: continuity of probabilities. ## Why limits of sets? So far, collections like $A_1,A_2,A_3,\ldots$ were mostly just indexed families. The order did not matter much when taking unions or intersections. From now we begin to care about the order. How does the sequence of sets evolve over time - we might be dealing with a time series, their index being time. Or they may evolve so that an events may depend on a previous one. In any case, we want to understand the long-term behavior of a sequence of sets. A sequence of sets is meant to evolve: $A_1, A_2, A_3, \ldots$ The question is whether this sequence has a limiting set. :::{.column-margin #fig-l08-slide-01} ![A sequence of sets may appear to approach a limiting set.](images/l08-s01.png){fig-alt="The slide introduces the question of defining the limit of a sequence of sets A n. It shows a sequence of circular sets moving or shrinking toward a possible limiting object." group="slides"} The intuitive picture of sets “approaching” a limit works well in metric spaces, where distance is meaningful. But measure theory also needs a definition for abstract sets, where geometric distance may not exist. ::: ## First example: nested intervals Let $\Omega=\mathbb{R},\quad A_n=\left[0,1-\frac{1}{n+1}\right].$ Then $$ A_1=\left[0,\frac12\right], \qquad A_2=\left[0,\frac23\right], \qquad A_3=\left[0,\frac34\right], \quad \ldots $$ The sets are increasing: $A_1\subseteq A_2\subseteq A_3\subseteq\cdots.$, intuitively, they approach $[0,1)$ :::{.column-margin #fig-l08-slide-02} ![Increasing closed intervals approaching the half-open interval $[0,1)$](images/l08-s02.png){group="slides"} The slide defines A n as the closed interval from 0 to 1 minus 1 over n plus 1. It lists A1 as [0,1/2], A2 as [0,2/3], A3 as [0,3/4], and suggests the sets approach $[0,1)$. The right endpoints move toward $1$ but never reach it. This suggests that the limiting set should include $0$ but exclude $1$. ::: This example is geometrically clear, but we need a definition that also works for arbitrary measurable sets. ## Warm-up: $\limsup$ and $\liminf$ of real sequences {#sec-limsup-liminf-numbers} Before defining limits of sets, recall the analogous ideas for a real sequence $a_1,a_2,a_3,\ldots.$ The **limit superior** or **limit supremum** tracks the eventual upper envelope of the sequence. :::{.column-margin #fig-l08-slide-03 group="slides"} ![A real sequence oscillating between upper and lower envelopes.](images/l08-s03.png){fig-alt="The slide shows a plotted real sequence with points A1, A2, A3 and so on, visually clustered between an upper and lower curve." group="slides"} The curves are only visual aids. The point is that a sequence may not settle immediately, but it may have an eventual upper and lower boundary. ::: For each $n$, look at the tail $\{a_m:m\geq n\}.$ Take the supremum of that tail: $\sup_{m\geq n} a_m.$ As $n$ increases, we discard more initial terms, so these tail suprema form a non-increasing sequence. The limit of these tail suprema is the $\limsup$: $$ \limsup_{n\to\infty} a_n = \lim_{n\to\infty} \sup_{m\geq n} a_m $$ {#eq-limsup-def-numbers} Equivalently, $$ \limsup_{n\to\infty} a_n = \inf_{n\geq 1}\sup_{m\geq n}a_m. $$ {#eq-limsup-numbers} :::{.column-margin #fig-l08-slide-04} ![The limsup is the limit of tail suprema.](images/l08-s04.png){fig-alt="The slide explains limsup for a sequence of real numbers by removing initial terms and recomputing the supremum of the remaining tail. The resulting tail suprema form a non-increasing sequence." group="slides"} The $\limsup$ is what remains of the sequence’s upper behavior after every finite prefix has been ignored. ::: Similarly, the **limit inferior** or **limit infimum** tracks the eventual lower envelope: $$ \liminf_{n\to\infty} a_n = \lim_{n\to\infty} \inf_{m\geq n}a_m = \sup_{n\geq 1}\inf_{m\geq n}a_m. $$ :::{.column-margin #fig-l08-slide-05} ![The liminf is the limit of tail infima.](images/l08-s05.png){fig-alt="The slide explains liminf by removing initial terms and recomputing the infimum of the remaining tail. The resulting tail infima form a non-decreasing sequence." group="slides"} The $\liminf$ is what remains of the sequence’s lower behavior after every finite prefix has been ignored. ::: If the two envelopes meet, the sequence has a limit: $$ \limsup_{n\to\infty} a_n = \liminf_{n\to\infty} a_n = L. $$ Then $$ \lim_{n\to\infty} a_n=L. $$ :::{.column-margin #fig-l08-slide-06} ![A real sequence has a limit when limsup and liminf agree.](images/l08-s06.png){fig-alt="The slide states that if the limsup and liminf of a real sequence are equal, their common value is the limit of the sequence." group="slides"} For numbers, the limit exists exactly when the eventual upper and lower envelopes collapse to the same value. ::: The set version will imitate this idea, but with unions and intersections replacing suprema and infima. ## Monotone sequences of sets ### Decreasing sets Suppose $$ A_1\supseteq A_2\supseteq A_3\supseteq\cdots. $$ Then the sets are decreasing. The natural limiting set is the part that remains forever: $$ A=\bigcap_{n=1}^{\infty}A_n. $$ We write $$ A_n\downarrow A. $$ :::{.column-margin #fig-l08-slide-07} ![A decreasing sequence of sets converges to the intersection.](images/l08-s07.png){fig-alt="The slide shows nested sets A1 containing A2 containing A3 and so on. It suggests that the limiting set is the intersection of all A n." group="slides"} For decreasing sets, the limit is the common core that survives every stage: $\bigcap_n A_n$. ::: ### Increasing sets Suppose $$ A_1\subseteq A_2\subseteq A_3\subseteq\cdots. $$ Then the sets are increasing. The natural limiting set is everything that eventually appears: $$ A=\bigcup_{n=1}^{\infty}A_n. $$ We write $$ A_n\uparrow A. $$ :::{.column-margin #fig-l08-slide-08} ![An increasing sequence of sets converges to the union.](images/l08-s08.png){fig-alt="The slide shows nested sets A1 inside A2 inside A3 and so on. It suggests that the limiting set is the union of all A n." group="slides"} For increasing sets, the limit is the accumulated set of everything that appears at some finite stage: $\bigcup_n A_n$. ::: These two cases motivate the general definitions. ## General sequences of sets Now let $$ A_1,A_2,A_3,\ldots $$ be any sequence of sets. They need not be increasing, decreasing, disjoint, or nested. To define the set-theoretic $\limsup$ and $\liminf$, use tails of the sequence. ### Tail unions For each $n$, define the tail union $$ \bigcup_{m=n}^{\infty}A_m. $$ As $n$ increases, we remove more initial sets, so the tail unions decrease: $$ \bigcup_{m=1}^{\infty}A_m \supseteq \bigcup_{m=2}^{\infty}A_m \supseteq \bigcup_{m=3}^{\infty}A_m \supseteq \cdots. $$ The limit of this decreasing sequence is its intersection. :::{.column-margin #fig-l08-slide-09} ![Tail unions form a decreasing sequence of sets.](images/l08-s09.png){fig-alt="The slide forms the tail unions of a general sequence of sets: the union of all A m from m equals 1 onward, then from m equals 2 onward, then from m equals 3 onward. These tail unions form a decreasing sequence." group="slides"} Even if the original $A_n$ are not monotone, the tail unions are monotone decreasing. ::: ### Tail intersections For each $n$, define the tail intersection $$ \bigcap_{m=n}^{\infty}A_m. $$ As $n$ increases, we remove more initial constraints, so the tail intersections increase: $$ \bigcap_{m=1}^{\infty}A_m \subseteq \bigcap_{m=2}^{\infty}A_m \subseteq \bigcap_{m=3}^{\infty}A_m \subseteq \cdots. $$ The limit of this increasing sequence is its union. :::{.column-margin #fig-l08-slide-10} ![Tail intersections form an increasing sequence of sets.](images/l08-s10.png){fig-alt="The slide forms the tail intersections of a general sequence of sets: the intersection of all A m from m equals 1 onward, then from m equals 2 onward, then from m equals 3 onward. These tail intersections form an increasing sequence." group="slides"} The tail intersections are monotone increasing because fewer sets are being intersected as the starting index moves forward. ::: ## Definitions: $\limsup$ and $\liminf$ of sets ::: {#def-limsup-sets} ## Limit superior of sets For a sequence of sets $A_1,A_2,\ldots$, define $$ \limsup_{n\to\infty} A_n = \bigcap_{n=1}^{\infty} \bigcup_{m=n}^{\infty}A_m. $$ ::: ::: {#def-liminf-sets} ## Limit inferior of sets For a sequence of sets $A_1,A_2,\ldots$, define $$ \liminf_{n\to\infty} A_n = \bigcup_{n=1}^{\infty} \bigcap_{m=n}^{\infty}A_m. $$ ::: :::{.column-margin #fig-l08-slide-11} ![Definitions of limsup and liminf for sequences of sets.](images/l08-s11.png){fig-alt="The slide defines limsup A n as the intersection over n of the union over m at least n of A m, and liminf A n as the union over n of the intersection over m at least n of A m." group="slides"} The $\limsup$ is built from decreasing tail unions. The $\liminf$ is built from increasing tail intersections. ::: Both are sets. They are made entirely from countable unions and intersections. ## Basic containment For every sequence of sets, $$ \liminf_{n\to\infty}A_n \subseteq \limsup_{n\to\infty}A_n. $$ :::{.column-margin #fig-l08-slide-12} ![The liminf set is contained in the limsup set.](images/l08-s12.png){fig-alt="The slide states that the liminf of a sequence of sets is contained in the limsup of the sequence, and begins a proof by taking an element omega in the liminf." group="slides"} The lower limiting set is always contained in the upper limiting set, just as $\liminf a_n\leq\limsup a_n$ for real sequences. ::: Proof sketch: Take $$ \omega\in\liminf A_n = \bigcup_{n=1}^{\infty}\bigcap_{m=n}^{\infty}A_m. $$ Then for some fixed $N$, $$ \omega\in\bigcap_{m=N}^{\infty}A_m. $$ So $\omega\in A_m$ for every $m\geq N$. Therefore, for every $n$, $\omega$ belongs to the tail union $$ \bigcup_{m=n}^{\infty}A_m. $$ Hence $$ \omega\in \bigcap_{n=1}^{\infty} \bigcup_{m=n}^{\infty}A_m = \limsup A_n. $$ :::{.column-margin #fig-l08-slide-13} ![Proof that liminf is contained in limsup.](images/l08-s13.png){fig-alt="The slide completes the proof that if omega is in the liminf, then omega is in all sufficiently late A m, hence belongs to every tail union and therefore to the limsup." group="slides"} If $\omega$ is eventually in all the $A_n$, then it is certainly in infinitely many of them. This proves $\liminf A_n\subseteq\limsup A_n$. ::: ## Definition of the limit of sets ::: {#def-limit-of-sets} ## Limit of a sequence of sets If $$ \liminf_{n\to\infty}A_n = \limsup_{n\to\infty}A_n, $$ then the common set is called the **limit** of the sequence, and we write $$ \lim_{n\to\infty}A_n = \liminf_{n\to\infty}A_n = \limsup_{n\to\infty}A_n. $$ ::: If the two sets are not equal, then the sequence of sets does not have a limit. :::{.column-margin #fig-l08-slide-14} ![A sequence of sets has a limit when liminf and limsup agree.](images/l08-s14.png){fig-alt="The slide defines the limit of a sequence of sets as the common value when the liminf and limsup are equal." group="slides"} The set limit exists exactly when the eventual lower and upper set behavior agree. ::: ## Checking the monotone cases ### Decreasing case Suppose $$ A_1\supseteq A_2\supseteq A_3\supseteq\cdots. $$ Then $$ \limsup A_n = \bigcap_{n=1}^{\infty}A_n, $$ because each tail union is simply its first set: $$ \bigcup_{m=n}^{\infty}A_m=A_n. $$ :::{.column-margin #fig-l08-slide-15} ![For decreasing sets, limsup is the intersection.](images/l08-s15.png){fig-alt="The slide computes limsup for a decreasing sequence of sets and shows that each tail union equals A n, so the limsup becomes the intersection of all A n." group="slides"} In a decreasing sequence, the largest set in each tail is the first one, so the tail union from $n$ onward is just $A_n$. ::: Similarly, $$ \liminf A_n = \bigcap_{n=1}^{\infty}A_n. $$ Thus, $$ A_n\downarrow A \quad\Rightarrow\quad \lim_n A_n=A=\bigcap_{n=1}^{\infty}A_n. $$ :::{.column-margin #fig-l08-slide-16} ![For decreasing sets, liminf is also the intersection.](images/l08-s16.png){fig-alt="The slide computes liminf for a decreasing sequence and shows it also equals the intersection of all A n. Therefore the limit exists and equals the intersection." group="slides"} For decreasing sequences, $\liminf$ and $\limsup$ agree, so the limit exists and equals the intersection. ::: ### Increasing case If $$ A_1\subseteq A_2\subseteq A_3\subseteq\cdots, $$ then the analogous result is $$ A_n\uparrow A \quad\Rightarrow\quad \lim_n A_n=A=\bigcup_{n=1}^{\infty}A_n. $$ :::{.column-margin #fig-l08-slide-17} ![For increasing sets, the limit is the union.](images/l08-s17.png){fig-alt="The slide asks the viewer to verify the increasing case: if A n increases, then limsup and liminf both equal the union of all A n." group="slides"} The increasing case is left as an exercise in the video: tail intersections become $A_n$, and tail unions all give the accumulated union. ::: ## Revisit the interval example Recall $$ A_n=\left[0,1-\frac{1}{n+1}\right]. $$ This is an increasing sequence, so the limit is the union: $$ \lim_{n\to\infty}A_n = \bigcup_{n=1}^{\infty} \left[0,1-\frac{1}{n+1}\right] = [0,1). $$ :::{.column-margin #fig-l08-slide-18} ![The interval example has limit [0,1).](images/l08-s18.png){fig-alt="The slide revisits A n equals [0, 1 minus 1 over n plus 1] and computes limsup and liminf directly, both giving the half-open interval [0,1)." group="slides"} The endpoint $1$ is approached but never included by any $A_n$, so it is not in the limiting set. ::: ## Alternating sets Consider a sequence alternating between two sets: $$ A_n= \begin{cases} B, & n \text{ odd},\\ C, & n \text{ even}. \end{cases} $$ Then every tail union contains both $B$ and $C$, so $$ \limsup A_n = B\cup C. $$ Every tail intersection contains only what is common to both, so $$ \liminf A_n = B\cap C. $$ Therefore the limit exists only if $$ B\cup C = B\cap C, $$ which happens when $B=C$. :::{.column-margin #fig-l08-slide-19} ![An alternating sequence has limsup B union C and liminf B intersect C.](images/l08-s19.png){fig-alt="The slide gives a coin-flipping-inspired example where A n alternates between two sets B and C. It computes limsup as B union C and liminf as B intersect C." group="slides"} Alternation usually prevents convergence. The upper limit sees points that appear repeatedly in either set, while the lower limit sees only points present in both. ::: ## Complements The complement of a $\limsup$ is a $\liminf$ of complements: $$ \left(\limsup_{n\to\infty}A_n\right)^c = \liminf_{n\to\infty}A_n^c. $$ Using the definition, $$ \left( \bigcap_{n=1}^{\infty} \bigcup_{m=n}^{\infty}A_m \right)^c = \bigcup_{n=1}^{\infty} \bigcap_{m=n}^{\infty}A_m^c. $$ This is just De Morgan’s law applied to countable intersections and unions. Likewise, $$ \left(\liminf_{n\to\infty}A_n\right)^c = \limsup_{n\to\infty}A_n^c. $$ :::{.column-margin #fig-l08-slide-20} ![Complements swap limsup and liminf.](images/l08-s20.png){fig-alt="The slide applies De Morgan's laws to show that the complement of limsup A n equals liminf of A n complement, and vice versa." group="slides"} Complements reverse the order of unions and intersections, so they swap $\limsup$ and $\liminf$. ::: ## Event language The $\limsup$ has a useful event interpretation. A point $\omega$ is in $$ \limsup A_n = \bigcap_{n=1}^{\infty}\bigcup_{m=n}^{\infty}A_m $$ exactly when, no matter how far out in the sequence we go, $\omega$ appears in some later $A_m$. So $\omega$ is in infinitely many of the $A_n$. This is often written $$ \limsup A_n = [A_n\ \text{i.o.}], $$ where `i.o.` means **infinitely often**. :::{.column-margin #fig-l08-slide-21} ![The limsup event means A n occurs infinitely often.](images/l08-s21.png){fig-alt="The slide explains that omega in limsup A n means omega belongs to infinitely many of the A n. It introduces the notation A n i.o., meaning infinitely often." group="slides"} In probability language, $\limsup A_n$ is the event that the events $A_n$ occur infinitely often. ::: The $\liminf$ means eventual membership: $$ \liminf A_n = \bigcup_{n=1}^{\infty}\bigcap_{m=n}^{\infty}A_m. $$ A point $\omega$ is in $\liminf A_n$ when there is some finite time $N$ after which $$ \omega\in A_m \qquad \text{for all } m\geq N. $$ That is, $\omega$ belongs to all but finitely many of the $A_n$. :::{.column-margin #fig-l08-slide-22} ![The liminf event means A n occurs eventually.](images/l08-s22.png){fig-alt="The slide explains that omega in liminf A n means omega is in all but finitely many of the A n. It mentions the notation A n eventually." group="slides"} The $\liminf$ event is stronger than the $\limsup$ event: eventually always implies infinitely often. ::: ## Measurability of set limits Suppose $(\Omega,\mathcal{F},P)$ is a probability space and $$ A_n\in\mathcal{F} \qquad \text{for all } n. $$ Because $\mathcal{F}$ is a $\sigma$-field, it is closed under countable unions and intersections. Therefore, $$ \limsup A_n = \bigcap_{n=1}^{\infty} \bigcup_{m=n}^{\infty}A_m \in\mathcal{F}, $$ and $$ \liminf A_n = \bigcup_{n=1}^{\infty} \bigcap_{m=n}^{\infty}A_m \in\mathcal{F}. $$ So it makes sense to discuss $$ P(\limsup A_n), \qquad P(\liminf A_n). $$ :::{.column-margin #fig-l08-slide-23} ![Limsup and liminf of measurable sets are measurable.](images/l08-s23.png){fig-alt="The slide states that if A n are in a sigma-field F, then limsup A n and liminf A n are also in F because they are built using countable unions and intersections." group="slides"} If each $A_n$ is measurable, then both limiting sets are measurable. This is exactly why $\sigma$-fields require countable closure. ::: If the limit exists, then $$ \lim_n A_n\in\mathcal{F}, $$ so $$ P\left(\lim_{n\to\infty}A_n\right) $$ is also well-defined. :::{.column-margin #fig-l08-slide-24} ![The probability of a set limit is well-defined when the limit exists.](images/l08-s24.png){fig-alt="The slide says that if liminf and limsup agree, then the limit set belongs to the sigma-field, so the probability of the limit set is defined." group="slides"} The limit of measurable sets, when it exists, is another measurable set. ::: ## Preview: continuity of probabilities The final question is whether probability commutes with limits of sets. Compare $$ P\left(\lim_{n\to\infty}A_n\right) $$ with $$ \lim_{n\to\infty}P(A_n). $$ These are different-looking operations: - first take a set limit, then measure it; - first measure each set, then take a numerical limit. The next lesson studies when these agree. This is called **continuity of probabilities**. :::{.column-margin #fig-l08-slide-25} ![Preview of continuity of probabilities.](images/l08-s25.png){fig-alt="The slide asks how P of the limit of A n relates to the limit of P of A n, previewing continuity of probabilities and the Borel-Cantelli lemmas." group="slides"} The next topic asks whether $P(\lim A_n)$ equals $\lim P(A_n)$. This connects set limits to ordinary limits of real numbers. ::: ## Takeaway The key definitions are: $$ \limsup_{n\to\infty}A_n = \bigcap_{n=1}^{\infty}\bigcup_{m=n}^{\infty}A_m, $$ $$ \liminf_{n\to\infty}A_n = \bigcup_{n=1}^{\infty}\bigcap_{m=n}^{\infty}A_m. $$ The interpretation is: $$ \limsup A_n = \{\omega:\omega\in A_n \text{ infinitely often}\}, $$ $$ \liminf A_n = \{\omega:\omega\in A_n \text{ eventually}\}. $$ The limit exists when these two sets are equal.