Limit Supremum & Limit Infimum

Author

John Robin Inston

Published

September 25, 2026

1 Review: Supremum and Infimum

For a full treatment see Supremum & Infimum. We work in the extended reals \(\overline{\mathbb{R}}=\mathbb{R}\cup\{\pm\infty\}\), where every subset has a supremum and an infimum.

Let \(A\subseteq\overline{\mathbb{R}}\). The supremum \(\sup A\) is the least upper bound of \(A\) and the infimum \(\inf A\) is the greatest lower bound of \(A\). Equivalently, \(s=\sup A\) if and only if \(a\leq s\) for all \(a\in A\) and, for every \(\epsilon>0\) (when \(s<\infty\)), there is \(a\in A\) with \(a>s-\epsilon\).

The supremum and infimum need not be attained, unlike the max and min. For example \(\sup(0,1)=1\notin(0,1)\). Also \(\inf A\leq\sup A\) for nonempty \(A\), and \(A\subseteq B\implies\sup A\leq\sup B\) and \(\inf A\geq\inf B\).

2 Limit Supremum & Infimum of a Sequence

Let \(\{x_n\}_{n\in\mathbb{N}}\subseteq\overline{\mathbb{R}}\). The sequence of tail suprema \(s_n:=\sup_{m\geq n}x_m\) is decreasing, since we take the supremum over fewer terms as \(n\) grows. Dually the tail infima \(i_n:=\inf_{m\geq n}x_m\) are increasing. Monotone sequences in \(\overline{\mathbb{R}}\) always have a limit, which motivates the following.

\[ \limsup_{n\to\infty}x_n:=\lim_{n\to\infty}\sup_{m\geq n}x_m=\inf_{n\geq1}\sup_{m\geq n}x_m, \qquad \liminf_{n\to\infty}x_n:=\lim_{n\to\infty}\inf_{m\geq n}x_m=\sup_{n\geq1}\inf_{m\geq n}x_m. \]

Both always exist in \(\overline{\mathbb{R}}\) (even when \(\lim x_n\) does not). For example, \(x_n=(-1)^n\) has \(\limsup x_n=1\) and \(\liminf x_n=-1\).

3 Limit Supremum & Infimum of a Sequence of Sets

Order subsets of \(X\) by inclusion. A least upper bound of \(\{A_n\}\) is the smallest set containing every \(A_n\), and a greatest lower bound is the largest set contained in every \(A_n\). Hence \[ \sup_{n\geq1}A_n=\bigcup_{n=1}^\infty A_n,\qquad\inf_{n\geq1}A_n=\bigcap_{n=1}^\infty A_n. \] Mimicking the definition for sequences, the tail supremum \(B_n=\bigcup_{m\geq n}A_m\) is decreasing and the tail infimum \(C_n=\bigcap_{m\geq n}A_m\) is increasing. A decreasing sequence of sets converges to its intersection and an increasing sequence to its union.

\[ \limsup_{n\to\infty}A_n:=\bigcap_{n=1}^\infty\bigcup_{m=n}^\infty A_m, \qquad \liminf_{n\to\infty}A_n:=\bigcup_{n=1}^\infty\bigcap_{m=n}^\infty A_m. \]

The two definitions are the same under the dictionary \(\sup\leftrightarrow\cup\), \(\inf\leftrightarrow\cap\), \(\leq\leftrightarrow\subseteq\). This is made precise by indicator functions.

\(\mathbf{1}_{\limsup A_n}=\limsup\mathbf{1}_{A_n}\) and \(\mathbf{1}_{\liminf A_n}=\liminf\mathbf{1}_{A_n}\) pointwise.

The proof follows from the characterisations in the next section: \(\mathbf{1}_{A_n}(x)\in\{0,1\}\), so its limsup is \(1\) exactly when it equals \(1\) infinitely often.

4 Equivalent Ways to Think About Limsup & Liminf

4.1 Sets: “infinitely often” and “eventually”

\[ \begin{align} \limsup_{n\to\infty}A_n&=\{x: x\in A_n\text{ for infinitely many }n\}=\{A_n\text{ i.o.}\},\\ \liminf_{n\to\infty}A_n&=\{x: x\in A_n\text{ for all but finitely many }n\}=\{A_n\text{ eventually}\}. \end{align} \]

Proof. \(x\in\limsup A_n\) iff for every \(n\) there is \(m\geq n\) with \(x\in A_m\), iff \(\{m: x\in A_m\}\) is unbounded, iff it is infinite. Also \(x\in\liminf A_n\) iff there is \(n\) with \(x\in A_m\) for all \(m\geq n\), iff \(\{m: x\notin A_m\}\) is bounded, iff it is finite.

So \(\limsup\) is the set of points that keep returning to the sets, whereas \(\liminf\) is the set of points that eventually stay in the sets forever. In probability, \(\limsup A_n\) is the event “\(A_n\) occurs infinitely often” and \(\liminf A_n\) is “\(A_n\) occurs eventually”, the starting point for the Borel-Cantelli Lemma.

4.2 Sequences: “frequently” and “eventually”

Let \(L\in\overline{\mathbb{R}}\). Then \(L=\limsup x_n\) if and only if

for every \(a>L\), \(x_n<a\) for all but finitely many \(n\) (eventually), and
for every \(a<L\), \(x_n>a\) for infinitely many \(n\) (frequently).

Dually, \(\ell=\liminf x_n\) if and only if for every \(a<\ell\), \(x_n>a\) eventually, and for every \(a>\ell\), \(x_n<a\) infinitely often.

Proof. (\(\Rightarrow\)) If \(a>L=\inf_ns_n\) there is \(N\) with \(s_N<a\), so \(x_m<a\) for all \(m\geq N\). If \(a<L\) then \(s_n\geq L>a\) for every \(n\), so for every \(n\) there is \(m\geq n\) with \(x_m>a\), hence \(x_m>a\) for infinitely many \(m\).

(\(\Leftarrow\)) Assume 1 and 2. If \(a>L\), 1 gives \(s_n\leq a\) eventually, so \(\limsup x_n\leq a\), hence \(\limsup x_n\leq L\). If \(a<L\), 2 gives \(s_n\geq a\) for all \(n\), so \(\limsup x_n\geq a\), hence \(\limsup x_n\geq L\).

The liminf statement follows by applying this to \(-x_n\) using the duality below.

4.3 Sequences: subsequential limits

\(\limsup x_n\) is the largest and \(\liminf x_n\) the smallest limit (in \(\overline{\mathbb{R}}\)) of a subsequence of \(\{x_n\}\).

Proof. If \(x_{n_k}\to\ell\) then \(x_{n_k}\leq s_{n_k}\) and \(s_{n_k}\to\limsup x_n\), so \(\ell\leq\limsup x_n\). Conversely, for \(L=\limsup x_n\) finite, characterisation 2 with \(a=L-1/k\), together with characterisation 1 with \(a=L+1/k\), lets us choose \(n_1<n_2<\cdots\) with \(L-1/k<x_{n_k}<L+1/k\), so \(x_{n_k}\to L\). The cases \(L=\pm\infty\) are similar. The liminf is analogous.

4.4 Summary

	\(\limsup\)	\(\liminf\)
Sequences	\(\inf_n\sup_{m\geq n}x_m\)	\(\sup_n\inf_{m\geq n}x_m\)
Sets	\(\bigcap_n\bigcup_{m\geq n}A_m\)	\(\bigcup_n\bigcap_{m\geq n}A_m\)
Meaning	frequently / infinitely often	eventually / all but finitely many
Subsequences	largest limit point	smallest limit point

5 Basic Inequalities and Identities

5.1 For sequences

\(\liminf x_n\leq\limsup x_n\).

Proof. For every \(n\), \(i_n=\inf_{m\geq n}x_m\leq x_n\leq\sup_{m\geq n}x_m=s_n\). Hence \(i_n\leq s_n\), and taking \(n\to\infty\) gives the claim.

\(\{x_n\}\) converges in \(\overline{\mathbb{R}}\) if and only if \(\liminf x_n=\limsup x_n\), in which case \(\lim x_n=\liminf x_n=\limsup x_n\).

Proof. (\(\Rightarrow\)) Let \(x_n\to x\). For finite \(x\) and \(\epsilon>0\) we have \(x-\epsilon<x_n<x+\epsilon\) eventually, so \(x-\epsilon\leq i_n\leq s_n\leq x+\epsilon\) eventually. Hence \(x-\epsilon\leq\liminf x_n\leq\limsup x_n\leq x+\epsilon\), and \(\epsilon\to0\) gives equality. If \(x=+\infty\), then for every \(M\) we have \(x_n>M\) eventually, so \(\liminf x_n\geq M\) for all \(M\), hence \(\liminf x_n=\limsup x_n=+\infty\) (similarly for \(-\infty\)).

(\(\Leftarrow\)) Let \(L=\liminf x_n=\limsup x_n\) and take \(a<L<b\) in \(\overline{\mathbb{R}}\). By the characterisation theorem, \(x_n<b\) eventually (applied to the limsup) and \(x_n>a\) eventually (applied to the liminf). So \(x_n\in(a,b)\) eventually, which is the definition of \(x_n\to L\) in the topology of \(\overline{\mathbb{R}}\).

For sequences in \(\overline{\mathbb{R}}\):

\(\limsup(-x_n)=-\liminf x_n\).
If \(x_n\leq y_n\) for all \(n\), then \(\limsup x_n\leq\limsup y_n\) and \(\liminf x_n\leq\liminf y_n\).
\(\limsup(x_n+y_n)\leq\limsup x_n+\limsup y_n\) and \(\liminf(x_n+y_n)\geq\liminf x_n+\liminf y_n\), whenever the right-hand sides are defined (no \(\infty-\infty\)).

Proof.

\(\sup_{m\geq n}(-x_m)=-\inf_{m\geq n}x_m\); take the limit. (2) Each tail supremum (resp. infimum) is monotone in the sequence, so pass to the limit. (3) For \(m\geq n\), \(x_m+y_m\leq\sup_{k\geq n}x_k+\sup_{k\geq n}y_k\), so \(\sup_{m\geq n}(x_m+y_m)\leq s_n^x+s_n^y\); let \(n\to\infty\). The liminf statement follows from (1) applied to the first.

Note: The inequality in (3) can be strict, e.g. \(x_n=(-1)^n\), \(y_n=-(-1)^n\) gives \(x_n+y_n=0\) but \(\limsup x_n+\limsup y_n=2\). If \(x_n\) converges then equality holds.

5.2 For sets

\(\liminf A_n\subseteq\limsup A_n\).
Duality: \((\limsup A_n)^c=\liminf A_n^c\) and \((\liminf A_n)^c=\limsup A_n^c\).
\(\limsup(A_n\cup B_n)=\limsup A_n\cup\limsup B_n\) and \(\liminf(A_n\cap B_n)=\liminf A_n\cap\liminf B_n\).
If \(A_n\uparrow A\) or \(A_n\downarrow A\) then \(\lim A_n\) exists and equals \(A\), i.e. \(\liminf A_n=\limsup A_n=A\).