Set-theoretic limit

In mathematics, the limit of a sequence of sets A₁, A₂, ... (subsets of a common set X) is a set whose elements are determined by the sequence in either of two equivalent ways: (1) by upper and lower bounds on the sequence that converge monotonically to the same set (analogous to convergence of real-valued sequences) and (2) by convergence of a sequence of indicator functions which are themselves real-valued. As is the case with sequences of other objects, convergence is not necessary or even usual.

More generally, again analogous to real-valued sequences, the less restrictive limit infimum and limit supremum of a set sequence always exist and can be used to determine convergence: the limit exists if the limit infimum and limit supremum are identical. (See below). Such set limits are essential in measure theory and probability.

It is a common misconception that the limits infimum and supremum described here involve sets of accumulation points, that is, sets of x = lim_k→∞x_k, where each x_k is in some A_nk. This is only true if convergence is determined by the discrete metric (that is, x_n → x iff there is N such that x_n = x for all n ≥ N). This article is restricted to that situation as it is the only one relevant for measure theory and probability. See the examples below. (On the other hand, there are more general topological notions of set convergence that do involve accumulation points under different metrics or topologies.)

Definitions

The two definitions

Suppose that $\{A_n\}_{n=1}^\infty$ is a sequence of sets. The two equivalent definitions are as follows.

• Using union and intersection, define

$\liminf_{n \rightarrow \infty} A_n = \bigcup_{n \ge 1} \bigcap_{j \geq n} A_j$

and

$\limsup_{n \rightarrow \infty} A_n = \bigcap_{n \ge 1} \bigcup_{j \geq n} A_j.$

If these two sets are equal, then the set-theoretic limit of the sequence A_n exists and is equal to that common set. Either set as described above can be used to get the limit, and there may be other means to get the limit as well.

• Using indicator functions, let 1_An(x) equal 1 if x is in A_n and 0 otherwise. Define

$\liminf_{n \rightarrow \infty} A_n = \{ x \in X: \liminf_{n \rightarrow \infty} \mathbf{1}_{A_n}(x) = 1 \}$

and

$\limsup_{n \rightarrow \infty} A_n = \{ x \in X: \limsup_{n \rightarrow \infty} \mathbf{1}_{A_n}(x) = 1 \},$

where the expressions inside the brackets on the right are, respectively, the limit infimum and limit supremum of the real-valued sequence 1_An(x). Again, if these two sets are equal, then the set-theoretic limit of the sequence A_n exists and is equal to that common set, and either set as described above can be used to get the limit.

To see the equivalence of the definitions, consider the limit infimum. The use of DeMorgan's rule below explains why this suffices for the limit supremum. Since indicator functions take only values 0 and 1, lim inf_n→∞ 1_An(x) = 1 if and only if 1_An(x) takes value 0 only finitely many times. Equivalently, $\scriptstyle x\ \in\ \bigcup_{n \ge 1}\ \bigcap_{j \ge n} A_n$ for some n if and only if there exists n such that the element is in A_m for every m ≥ n, which is to say if and only if x ∉ A_n for only finitely many n.

Therefore, x is in the lim inf_n→∞ A_n iff x is in all except finitely many A_n. For this reason, a shorthand phrase for the limit infimum is "x ∈ A_n all except finitely often" (or "x ∈ A_n all but finitely often"), typically expressed by "A_n a.e.f.o." (or by "A_n a.b.f.o.").

Similarly, an element of X is in the limit supremum if, no matter how large n is there exists m ≥ n such that the element is in A_m. That is, x is in the limit supremum iff x is in infinitely many A_n. For this reason, a shorthand phrase for the limit supremum is "x ∈ A_n infinitely often", typically expressed by "A_n i.o.".

Monotone sequences

The sequence A_n is said to be nonincreasing if each A_n+1 ⊂ A_n and nondecreasing if each A_n ⊂ A_n+1. In each of these cases the set limit exists. Consider, for example, a nonincreasing sequence A_n. Then

$\bigcap_{j \geq n} A_j = \bigcap_{j \geq 1} A_j \text{ and } \bigcup_{j \geq n} A_j = A_n.$

From these it follows that

$\liminf_{n \rightarrow \infty} A_n = \bigcup_{n \geq 1} \bigcap_{j \geq n} A_j = \bigcap_{j \geq 1} A_j = \bigcap_{n \geq 1} \bigcup_{j \geq n} A_j = \limsup_{n \rightarrow \infty} A_n.$

Similarly, if A_n is nondecreasing then

$\lim_{n \rightarrow \infty} A_n = \bigcup_{j \geq 1} A_j.$

Properties

• If the limit of 1_An(x), as n goes to infinity, exists for all x then

$\lim_{n \rightarrow \infty} A_n = \{ x \in X: \lim_{i \rightarrow \infty} \mathbf{1}_{A_n}(x) = 1 \}.$

Otherwise, the limit for {A_n} does not exist.

• It can be shown that the limit infimum is contained in the limit supremum:

$\liminf_{n\to\infty}A_n\subset\limsup_{n\to\infty}A_n,$

for example, simply by observing that x ∈ A_n all except finitely often implies x ∈ A_n infinitely often.

• Using the monotonicity of $\scriptstyle B_n = \bigcap_{j\ge n}A_j$ and of $\scriptstyle C_n = \bigcup_{j\ge n}A_j$,

$\liminf_{n\to\infty}A_n = \lim_{n\to\infty}\bigcap_{j\ge n}A_j\text{ and }\limsup_{n\to\infty}A_n = \lim_{n\to\infty}\bigcup_{j\ge n}A_j.$

• By using DeMorgan's rule twice, with set complement A^c = X\A,

$\liminf_{n \rightarrow \infty} A_n = \bigcup_i \Bigl(\bigcup_{j \geq n} A_j^c\Bigr)^c = \Bigl(\bigcap_n \bigcup_{j \geq n} A_j^c\Bigr)^c = \Bigl(\limsup_{n \rightarrow \infty} A_n^c\Bigr)^c.$

That is, x ∈ A_n all except finitely often is the same as x ∉ A_n finitely often.

• From the second definition above and the definitions for limit infimum and limit supremum of a real-valued sequence,

$\textbf{1}_{\liminf_{n \rightarrow \infty} A_n}(x) = \liminf_{n \rightarrow \infty}\textbf{1}_{A_j}(x) = \sup_{n \ge 1}\inf_{j \ge n}\textbf{1}_{A_j}(x)$

and

$\textbf{1}_{\limsup_{n \rightarrow \infty} A_n}(x) = \limsup_{n \rightarrow \infty}\textbf{1}_{A_j}(x) = \inf_{n \ge 1}\sup_{j \ge n}\textbf{1}_{A_j}(x).$

• Suppose $\scriptstyle\mathcal{F}$ is a σ-algebra of subsets of X. That is, $\scriptstyle\mathcal{F}$ is nonempty and is closed under complement and under unions and intersections of countably many sets. Then, by the first definition above, if each A_n ∈ $\scriptstyle\mathcal{F}$ then both lim inf_{n → ∞} A_n and lim sup_{n → ∞} A_n are elements of $\scriptstyle\mathcal{F}$.

Examples

• Let A_n = (−1/n, 1 − 1/n]. Then

$\liminf_{n \rightarrow \infty} A_n = \bigcup_n \bigcap_{j\ge n}\Bigl(-\frac{1}{j}, 1-\frac{1}{j}\Bigr] = \bigcup_n \Bigl[0, 1-\frac{1}{n}\Bigr] = [0, 1)$

and

$\limsup_{n \rightarrow \infty} A_n = \bigcap_n \bigcup_{j\ge n}\Bigl(-\frac{1}{j}, 1-\frac{1}{j}\Bigr] = \bigcap_n \Bigl(-\frac{1}{n}, 1\Bigr) = [0, 1).$

So lim_n→∞ A_n = (0, 1] exists.

• Change the previous example to A_n = ((−1)ⁿ/n, 1 − (−1)ⁿ/n]. Then

$\liminf_{n \rightarrow \infty} A_n = \bigcup_n \bigcap_{j\ge n} \Bigl(\frac{(-1)^j}{j}, 1-\frac{(-1)^j}{j}\Bigr] = \bigcup_n \Bigl(\frac{1}{2n}, 1-\frac{1}{2n}\Bigr] = (0, 1)$

and

$\limsup_{n \rightarrow \infty} A_n = \bigcap_n \bigcup_{j\ge n} \Bigl(\frac{(-1)^j}{j}, 1-\frac{(-1)^j}{j}\Bigr] = \bigcap_n \Bigl(-\frac{1}{2n-1}, 1+\frac{1}{2n-1}\Bigr] = [0, 1].$

So lim_n→∞A_n does not exist, despite the fact that the left and right endpoints of the intervals converge to 0 and 1, respectively.

• Let A_n = {0, 1/n, 2/n, ..., (n−1)/n, 1}. Then

$\bigcup_{j \geq n} A_j = \mathbb{Q}\cap[0,1]$

(which is all rational numbers between 0 and 1, inclusive) since even for j < n and 0 ≤ k ≤ j, k/j = (n×k)/(n×j) is an element of the above. Therefore,

$\limsup_{n \rightarrow \infty} A_n = \mathbb{Q}\cap[0,1].$

On the other hand,

$\bigcap_{j \geq n} A_j = \{0,1\},$

which implies

$\liminf_{n \rightarrow \infty} A_n = \{0,1\}.$

In this case, the sequence A₁, A₂, ... does not have a limit. Note that lim sup_n→∞ A_n is not the set of accumulation points, which would be the entire interval [0, 1] (according to the usual Euclidean metric).

Probability uses

Set limits, particularly the limit infimum and the limit supremum, are essential for probability and measure theory. Such limits are used to calculate (or prove) the probabilities and measures of other, more purposeful, sets. For the following, $\scriptstyle(X,\mathcal{F},\mathbb{P})$ is a probability space, which means $\scriptstyle\mathcal{F}$ is a σ-algebra of subsets of $\scriptstyle X$ and $\scriptstyle\mathbb{P}$ is a probability measure defined on that σ-algebra. Sets in the σ-algebra are known as events.

If A₁, A₂, ... is a sequence of events in $\scriptstyle\mathcal{F}$ and lim_n→∞ A_n exists then

$\mathbb{P}(\lim_{n \rightarrow \infty} A_n) = \lim_{n \rightarrow \infty} \mathbb{P}(A_n).$

Borel–Cantelli lemmas

In probability, the two Borel–Cantelli lemmas can be useful for showing that the limsup of a sequence of events has probability equal to 1 or to 0. The statement of the first (original) Borel–Cantelli lemma is

$\text{if } \sum_{n=1}^{\infty} \mathbb{P}(A_n) < \infty \text{ then } \mathbb{P}(\limsup_{n \rightarrow \infty} A_n) = 0.$

The second Borel–Cantelli lemma is a partial converse:

$\text{if } A_1, A_2, \dots \text{ are independent events and } \sum_{n=1}^\infty \mathbb{P}(A_n) = \infty \text{ then } \mathbb{P}(\limsup_{n \rightarrow \infty} A_n) = 1.$

Almost sure convergence

One of the most important applications to probability is for demonstrating the almost sure convergence of a sequence of random variables. The event that a sequence of random variables Y₁, Y₂, ... converges to another random variable Y is formally expressed as $\scriptstyle\{\limsup_{n\to\infty}|Y_n - Y|=0\}$. It would be a mistake, however, to write this simply as a limsup of events. Instead, the complement of the event is

$\{\limsup_{n\to\infty}|Y_n - Y| \neq 0\} = \{\limsup_{n\to\infty}|Y_n - Y| > \frac{1}{k} \text{ for some } k\}$

$= \bigcup_{k\ge 1}\bigcap_{n\ge 1}\bigcup_{j \ge n} \{|Y_n - Y| > \frac{1}{k}\} = \lim_{k\to\infty}\limsup_{n\to\infty} \{|Y_n - Y| > \frac{1}{k}\}.$

Therefore,

$\mathbb{P}(\{\limsup_{n\to\infty}|Y_n - Y| \neq 0\}) = \lim_{k\to\infty} \mathbb{P}(\limsup_{n\to\infty} \{|Y_n - Y| > \frac{1}{k}\}).$