In mathematics, the limit of a sequence of sets ${\displaystyle A_{1},A_{2},\ldots }$ (subsets of a common set ${\displaystyle X}$) 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.

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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 ${\displaystyle x=\lim _{k\to \infty }x_{k},}$ where each ${\displaystyle x_{k}}$ is in some ${\displaystyle A_{n_{k}}.}$ This is only true if convergence is determined by the discrete metric (that is, ${\displaystyle x_{n}\to x}$ if there is ${\displaystyle N}$ such that ${\displaystyle x_{n}=x}$ for all ${\displaystyle n\geq N}$). 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.)

• If the limit of ${\displaystyle \mathbb {1} _{A_{n}}(x),}$ as ${\displaystyle n}$ goes to infinity, exists for all ${\displaystyle x}$ then
  $${\displaystyle \lim _{n\to \infty }A_{n}=\left\{x\in X:\lim _{n\to \infty }\mathbb {1} _{A_{n}}(x)=1\right\}.}$$

  
  Otherwise, the limit for ${\displaystyle \left(A_{n}\right)}$ does not exist.
• It can be shown that the limit infimum is contained in the limit supremum:
  $${\displaystyle \liminf _{n\to \infty }A_{n}\subseteq \limsup _{n\to \infty }A_{n},}$$

  
  for example, simply by observing that ${\displaystyle x\in A_{n}}$ all but finitely often implies ${\displaystyle x\in A_{n}}$ infinitely often.
• Using the monotonicity of ${\textstyle B_{n}=\bigcap _{j\geq n}A_{j}}$ and of ${\textstyle C_{n}=\bigcup _{j\geq n}A_{j},}$
  $${\displaystyle \liminf _{n\to \infty }A_{n}=\lim _{n\to \infty }\bigcap _{j\geq n}A_{j}\quad {\text{ and }}\quad \limsup _{n\to \infty }A_{n}=\lim _{n\to \infty }\bigcup _{j\geq n}A_{j}.}$$

  
• By using De Morgan's law twice, with set complement ${\displaystyle A^{c}:=X\setminus A,}$
  $${\displaystyle \liminf _{n\to \infty }A_{n}=\bigcup _{n}\left(\bigcup _{j\geq n}A_{j}^{c}\right)^{c}=\left(\bigcap _{n}\bigcup _{j\geq n}A_{j}^{c}\right)^{c}=\left(\limsup _{n\to \infty }A_{n}^{c}\right)^{c}.}$$

  
  That is, ${\displaystyle x\in A_{n}}$ all but finitely often is the same as ${\displaystyle x\not \in A_{n}}$ finitely often.
• From the second definition above and the definitions for limit infimum and limit supremum of a real-valued sequence,
  $${\displaystyle \mathbb {1} _{\liminf _{n\to \infty }A_{n}}(x)=\liminf _{n\to \infty }\mathbb {1} _{A_{n}}(x)=\sup _{n\geq 1}\inf _{j\geq n}\mathbb {1} _{A_{j}}(x)}$$

  
  and
  $${\displaystyle \mathbb {1} _{\limsup _{n\to \infty }A_{n}}(x)=\limsup _{n\to \infty }\mathbb {1} _{A_{n}}(x)=\inf _{n\geq 1}\sup _{j\geq n}\mathbb {1} _{A_{j}}(x).}$$

  
• Suppose ${\displaystyle {\mathcal {F}}}$ is a 𝜎-algebra of subsets of ${\displaystyle X.}$ That is, ${\displaystyle {\mathcal {F}}}$ is nonempty and is closed under complement and under unions and intersections of countably many sets. Then, by the first definition above, if each ${\displaystyle A_{n}\in {\mathcal {F}}}$ then both ${\displaystyle \liminf _{n\to \infty }A_{n}}$ and ${\displaystyle \limsup _{n\to \infty }A_{n}}$ are elements of ${\displaystyle {\mathcal {F}}.}$

• Let ${\displaystyle A_{n}=\left(-{\tfrac {1}{n}},1-{\tfrac {1}{n}}\right].}$ Then

$${\displaystyle \liminf _{n\to \infty }A_{n}=\bigcup _{n}\bigcap _{j\geq n}\left(-{\tfrac {1}{j}},1-{\tfrac {1}{j}}\right]=\bigcup _{n}\left[0,1-{\tfrac {1}{n}}\right]=[0,1)}$$

and

$${\displaystyle \limsup _{n\to \infty }A_{n}=\bigcap _{n}\bigcup _{j\geq n}\left(-{\tfrac {1}{j}},1-{\tfrac {1}{j}}\right]=\bigcap _{n}\left(-{\tfrac {1}{n}},1\right)=[0,1)}$$

so ${\displaystyle \lim _{n\to \infty }A_{n}=[0,1)}$ exists.

• Change the previous example to ${\displaystyle A_{n}=\left({\tfrac {(-1)^{n}}{n}},1-{\tfrac {(-1)^{n}}{n}}\right].}$ Then

$${\displaystyle \liminf _{n\to \infty }A_{n}=\bigcup _{n}\bigcap _{j\geq n}\left({\tfrac {(-1)^{j}}{j}},1-{\tfrac {(-1)^{j}}{j}}\right]=\bigcup _{n}\left({\tfrac {1}{2n}},1-{\tfrac {1}{2n}}\right]=(0,1)}$$

and

$${\displaystyle \limsup _{n\to \infty }A_{n}=\bigcap _{n}\bigcup _{j\geq n}\left({\tfrac {(-1)^{j}}{j}},1-{\tfrac {(-1)^{j}}{j}}\right]=\bigcap _{n}\left(-{\tfrac {1}{2n-1}},1+{\tfrac {1}{2n-1}}\right]=[0,1],}$$

so ${\displaystyle \lim _{n\to \infty }A_{n}}$ does not exist, despite the fact that the left and right endpoints of the intervals converge to 0 and 1, respectively.

• Let ${\displaystyle A_{n}=\left\{0,{\tfrac {1}{n}},{\tfrac {2}{n}},\ldots ,{\tfrac {n-1}{n}},1\right\}.}$ Then

$${\displaystyle \bigcup _{j\geq n}A_{j}=\mathbb {Q} \cap [0,1]}$$

is the set of all rational numbers between 0 and 1 (inclusive), since even for ${\displaystyle j<n}$ and ${\displaystyle 0\leq k\leq j,}$ ${\displaystyle {\tfrac {k}{j}}={\tfrac {nk}{nj}}}$ is an element of the above. Therefore,

$${\displaystyle \limsup _{n\to \infty }A_{n}=\mathbb {Q} \cap [0,1].}$$

On the other hand,

$${\displaystyle \bigcap _{j\geq n}A_{j}=\{0,1\},}$$

which implies

$${\displaystyle \liminf _{n\to \infty }A_{n}=\{0,1\}.}$$

In this case, the sequence ${\displaystyle A_{1},A_{2},\ldots }$ does not have a limit. Note that ${\displaystyle \lim _{n\to \infty }A_{n}}$ is not the set of accumulation points, which would be the entire interval ${\displaystyle [0,1]}$ (according to the usual Euclidean metric).

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, ${\displaystyle (X,{\mathcal {F}},\mathbb {P} )}$ is a probability space, which means ${\displaystyle {\mathcal {F}}}$ is a σ-algebra of subsets of ${\displaystyle X}$ and ${\displaystyle \mathbb {P} }$ is a probability measure defined on that σ-algebra. Sets in the σ-algebra are known as events.

If ${\displaystyle A_{1},A_{2},\ldots }$ is a monotone sequence of events in ${\displaystyle {\mathcal {F}}}$ then ${\displaystyle \lim _{n\to \infty }A_{n}}$ exists and

$${\displaystyle \mathbb {P} \left(\lim _{n\to \infty }A_{n}\right)=\lim _{n\to \infty }\mathbb {P} \left(A_{n}\right).}$$

Borel–Cantelli lemmas

Main article: Borel–Cantelli lemma

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

First Borel–Cantelli lemma — If

$${\displaystyle \sum _{n=1}^{\infty }\mathbb {P} \left(A_{n}\right)<\infty }$$

then

$${\displaystyle \mathbb {P} \left(\limsup _{n\to \infty }A_{n}\right)=0.}$$

The second Borel–Cantelli lemma is a partial converse:

Second Borel–Cantelli lemma — If

$${\displaystyle A_{1},A_{2},\ldots }$$

are independent events and

$${\displaystyle \sum _{n=1}^{\infty }\mathbb {P} \left(A_{n}\right)=\infty }$$

then

$${\displaystyle \mathbb {P} \left(\limsup _{n\to \infty }A_{n}\right)=1.}$$

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 ${\displaystyle Y_{1},Y_{2},\ldots }$ converges to another random variable ${\displaystyle Y}$ is formally expressed as ${\textstyle \left\{\limsup _{n\to \infty }\left|Y_{n}-Y\right|=0\right\}.}$ It would be a mistake, however, to write this simply as a limsup of events. That is, this is not the event ${\textstyle \limsup _{n\to \infty }\left\{\left|Y_{n}-Y\right|=0\right\}}$! Instead, the complement of the event is

$${\displaystyle {\begin{aligned}\left\{\limsup _{n\to \infty }\left|Y_{n}-Y\right|\neq 0\right\}&=\left\{\limsup _{n\to \infty }\left|Y_{n}-Y\right|>{\frac {1}{k}}{\text{ for some }}k\right\}\\&=\bigcup _{k\geq 1}\bigcap _{n\geq 1}\bigcup _{j\geq n}\left\{\left|Y_{j}-Y\right|>{\tfrac {1}{k}}\right\}\\&=\lim _{k\to \infty }\limsup _{n\to \infty }\left\{\left|Y_{n}-Y\right|>{\tfrac {1}{k}}\right\}.\end{aligned}}}$$

Therefore,

$${\displaystyle \mathbb {P} \left(\left\{\limsup _{n\to \infty }\left|Y_{n}-Y\right|\neq 0\right\}\right)=\lim _{k\to \infty }\mathbb {P} \left(\limsup _{n\to \infty }\left\{\left|Y_{n}-Y\right|>{\tfrac {1}{k}}\right\}\right).}$$

• List of set identities and relations – Equalities for combinations of sets
• Set theory – Branch of mathematics that studies sets