Limsup, Liminf and Measure

Author

John Robin Inston

Published

October 11, 2026

1 Limsup, Liminf and Measure

For events \(A_n\) in a probability space, \(\mathbb{P}(\limsup A_n)\) is the probability that \(A_n\) occurs infinitely often. This note relates the limsup/liminf of the measures \(\mu(A_n)\) to the measure of the limsup/liminf of the sets \(A_n\). We recall the definitions from Limit Supremum & Limit Infimum, \[ \limsup_{n\to\infty}A_n=\bigcap_{n=1}^\infty\bigcup_{m=n}^\infty A_m, \qquad \liminf_{n\to\infty}A_n=\bigcup_{n=1}^\infty\bigcap_{m=n}^\infty A_m, \] and use continuity from above and below of measures, see Measure Spaces.

2 Inequalities

Let \((X,\mathcal{M},\mu)\) be a measure space and \(A_n\in\mathcal{M}\).

  1. \(\mu(\liminf A_n)\leq\liminf\mu(A_n)\).
  2. If \(\mu\left(\bigcup_{m\geq N}A_m\right)<\infty\) for some \(N\) (e.g. \(\mu\) is finite or a probability measure), then \(\limsup\mu(A_n)\leq\mu(\limsup A_n)\).

In particular for a probability measure \[ \mathbb{P}\left(\liminf_{n\to\infty}A_n\right)\leq\liminf_{n\to\infty}\mathbb{P}(A_n)\leq\limsup_{n\to\infty}\mathbb{P}(A_n)\leq\mathbb{P}\left(\limsup_{n\to\infty}A_n\right). \]

Proof.

  1. Let \(C_n=\bigcap_{m\geq n}A_m\), so \(C_n\uparrow\liminf A_n\). By continuity from below and since \(C_n\subseteq A_n\), \[ \mu(\liminf A_n)=\lim_{n\to\infty}\mu(C_n)\leq\liminf_{n\to\infty}\mu(A_n). \]

  2. Let \(B_n=\bigcup_{m\geq n}A_m\), so \(B_n\downarrow\limsup A_n\) and \(\mu(B_N)<\infty\). For \(m\geq n\) we have \(A_m\subseteq B_n\), so \(\mu(B_n)\geq\sup_{m\geq n}\mu(A_m)\). By continuity from above, \[ \mu(\limsup A_n)=\lim_{n\to\infty}\mu(B_n)\geq\lim_{n\to\infty}\sup_{m\geq n}\mu(A_m)=\limsup_{n\to\infty}\mu(A_n). \]

The middle inequality in the probabilistic statement is \(\liminf\leq\limsup\) for the sequence \(\mathbb{P}(A_n)\).

Part (1) is the set version of Fatou Lemma and part (2) is a reverse version.

If \(\mu\) is finite and \(\mu(A_n)\geq c>0\) for all \(n\), then \(\mu(\limsup A_n)\geq c\).

Proof. By part (2), \(\mu(\limsup A_n)\geq\limsup\mu(A_n)\geq c\).

3 Borel-Cantelli Lemmas

If \(\sum_{n}\mu(A_n)<\infty\) then \(\mu(\limsup A_n)=0\).

Proof. For every \(n\), \(\limsup A_n\subseteq\bigcup_{m\geq n}A_m\), so by monotonicity and countable subadditivity \[ \mu(\limsup A_n)\leq\sum_{m\geq n}\mu(A_m)\to0\quad(n\to\infty), \] as the tails of a convergent series tend to zero.

If \(A_n\) are independent events with \(\sum_n\mathbb{P}(A_n)=\infty\), then \(\mathbb{P}(\limsup A_n)=1\).

Proof. By De Morgan’s laws, \((\limsup A_n)^c=\bigcup_n\bigcap_{m\geq n}A_m^c\). Since the \(A_m^c\) are independent and \(1-p\leq e^{-p}\), for any \(N\geq n\) \[ \mathbb{P}\left(\bigcap_{m=n}^NA_m^c\right)=\prod_{m=n}^N(1-\mathbb{P}(A_m))\leq\exp\left(-\sum_{m=n}^N\mathbb{P}(A_m)\right)\to0\quad(N\to\infty). \] By continuity from above, \(\mathbb{P}\left(\bigcap_{m\geq n}A_m^c\right)=0\) for every \(n\), so by countable subadditivity \(\mathbb{P}((\limsup A_n)^c)=0\).

See also Borel-Cantelli Lemma.

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