Limsup and Liminf of Measurable Sequences

Tail Behavior and Oscillation

A sequence need not converge to have a useful description of its long-term behavior. Its terms may oscillate, or some values may recur arbitrarily far out in the sequence. The limsup records the largest values that persist arbitrarily far into the tail; the liminf records the corresponding smallest values. For a sequence of functions, these quantities are defined point by point and can themselves be measurable functions.

Let $(X,\mathcal{M},\mu)$ be any measure space; no finiteness, sigma-finiteness, or completeness assumption is needed. Suppose each $f_n:X\to\mathbb{R}$ is real-valued and $\mathcal{M}$-measurable. The definitions below are pointwise at every $x\in X$, not merely almost everywhere. Even though each $f_n$ is real-valued, the tail suprema, tail infima, limsup, and liminf may equal $+\infty$ or $-\infty$.

Definition: For $n\ge1$, the $n$th tail supremum and tail infimum at $x$ are $$ S_n(x)=\sup_{k\ge n}f_k(x),\qquad T_n(x)=\inf_{k\ge n}f_k(x). $$ The pointwise limsup and liminf are the extended-real values $$ \limsup_{n\to\infty}f_n(x)=\inf_{n\ge1}S_n(x),\qquad \liminf_{n\to\infty}f_n(x)=\sup_{n\ge1}T_n(x). $$ We abbreviate these as $L(x)$ and $I(x)$. A value occurs infinitely often in a set of indices if it occurs for arbitrarily large indices; a property holds eventually if it holds for all indices after some fixed index.

The tail suprema $S_n(x)$ decrease as $n$ increases, while the tail infima $T_n(x)$ increase. For every pair of indices $n,m$, choose any $k\ge\max(n,m)$. Then $T_n(x)\le f_k(x)\le S_m(x)$. Taking the supremum over $n$, and then the infimum over $m$, gives $I(x)\le L(x)$. Thus the liminf never exceeds the limsup.

Measurability: Strict Level Sets Matter

For a real number $a$, measurability can be checked by expressing strict level sets through countable unions and intersections of measurable sets. Rational numbers provide the strict margin needed in these expressions. In particular, “every term is below $a$” does not imply that a supremum is below $a$; the supremum might equal $a$.

Theorem: Under the assumptions above, $L$ and $I$ have measurable strict level sets: for every real $a$, each of $\{L<a\},\{L>a\},\{I<a\},\{I>a\}$ belongs to $\mathcal{M}$. The values $L$ and $I$ are allowed to be extended real.

Proof. We first establish four pointwise identities. For $L<a$, the definition $L(x)=\inf_n S_n(x)$ implies that some $S_n(x)<a$. Choose a rational $q$ with $S_n(x)<q<a$. Then $f_k(x)<q$ for every $k\ge n$. Conversely, if all these tail terms are less than some $q<a$, then $S_n(x)\le q<a$, so $L(x)<a$. Therefore

\{L<a\}= \bigcup_{\substack{q\in\mathbb{Q}\\q<a}} \ \bigcup_{n=1}^{\infty}\ \bigcap_{k=n}^{\infty}\{f_k<q\}.

Next, if $L(x)>a$, choose a rational $q$ with $a<q<L(x)$. Every tail has supremum at least $L(x)>q$, so every tail contains a term greater than $q$. Conversely, if every tail contains a term greater than some $q>a$, then every $S_n(x)\ge q$, hence $L(x)\ge q>a$. Thus

\{L>a\}= \bigcup_{\substack{q\in\mathbb{Q}\\q>a}} \ \bigcap_{n=1}^{\infty}\ \bigcup_{k=n}^{\infty}\{f_k>q\}.

For $I(x)>a$, choose a rational $q$ with $a<q<I(x)$. Since $I=\sup_nT_n$, some $T_n(x)>q$; every term in that tail is then greater than $q$. Conversely, if all terms from some index onward exceed $q>a$, then that tail infimum is at least $q$, giving $I(x)\ge q>a$. Hence

\{I>a\}= \bigcup_{\substack{q\in\mathbb{Q}\\q>a}} \ \bigcup_{n=1}^{\infty}\ \bigcap_{k=n}^{\infty}\{f_k>q\}.

Finally, if $I(x)<a$, choose a rational $q$ with $I(x)<q<a$. Every tail infimum is at most $I(x)$, so each tail contains a term less than $q$. Conversely, if every tail contains a term less than some $q<a$, then $T_n(x)\le q$ for every $n$, and $I(x)\le q<a$. Thus

\{I<a\}= \bigcup_{\substack{q\in\mathbb{Q}\\q<a}} \ \bigcap_{n=1}^{\infty}\ \bigcup_{k=n}^{\infty}\{f_k<q\}.

Because each $f_k$ is real-valued and measurable, its strict level sets $\{f_k<q\}$ and $\{f_k>q\}$ are in $\mathcal{M}$. Each right-hand side uses only countable unions and intersections, so all four sets are measurable. This proves the theorem. $\square$

The rational threshold is essential to the strict inequalities. For instance, $1-1/k<1$ for every $k$, but $\sup_{k\ge1}(1-1/k)=1$. Likewise $1/k>0$ for every $k$, but $\inf_{k\ge1}1/k=0$. The strict level-set identities above avoid both errors.

Worked Examples

Worked Example: An Oscillating Numerical Sequence

Let $a_n=(-1)^n+1/n$, for $n\ge1$. Find its limsup and liminf.

For even $n$, $a_n=1+1/n$, which approaches $1$ along the even indices. For odd $n$, $a_n=-1+1/n$, which approaches $-1$ along the odd indices. To verify that no other tail behavior changes these values, fix $N$. Every $k\ge N$ satisfies $a_k\le1+1/N$, since $(-1)^k\le1$ and $1/k\le1/N$. There are even indices $k\ge N$ with $1/k$ arbitrarily close to $0$, so the supremum of the $N$th tail is at least $1$. Consequently $1\le S_N\le1+1/N$, and $\inf_NS_N=1$.

Similarly, $a_k\ge-1$ for every $k$, while odd indices $k\ge N$ give values $-1+1/k$ arbitrarily close to $-1$. Thus $-1\le T_N\le-1+1/N$, and $\sup_NT_N=-1$. Therefore

\limsup_{n\to\infty}a_n=1,\qquad \liminf_{n\to\infty}a_n=-1.

Check: the first four terms are $0,\frac32,-\frac23,\frac54$. The even terms approach $1$ from above, and the odd terms approach $-1$ from above, consistent with the tail bounds.

Worked Example: A Measurable Sequence That Oscillates by Position

On $(\mathbb{R},\mathcal{L},m)$, let $f_n(x)=(-1)^n x$. Each $f_n$ is real-valued and Lebesgue measurable. Determine its pointwise limsup and liminf.

If $x\ge0$, even indices give $f_n(x)=x$ and odd indices give $f_n(x)=-x$. Both values occur in every tail, so its tail supremum is $x$ and its tail infimum is $-x$. If $x<0$, even indices give $x$, while odd indices give $-x>x$; the tail supremum is $-x$ and the tail infimum is $x$. At $x=0$, every term is zero. Combining these cases gives

\limsup_{n\to\infty}f_n(x)=|x|,\qquad \liminf_{n\to\infty}f_n(x)=-|x|\quad\text{for every }x\in\mathbb{R}.

Check: at $x=-2$, the even terms equal $-2$ and the odd terms equal $2$, so the limsup is $2=|-2|$ and the liminf is $-2=-|-2|$. At $x=0$, both formulas give zero. The resulting functions are measurable because the absolute-value function is continuous.

Worked Example: A Convergent Function Sequence

On $([0,1],\mathcal{B}_{[0,1]},m)$, where $m$ is Lebesgue measure restricted to the interval, let $f_n(x)=x^n$. Compute the limsup and liminf at every point.

At $x=0$, $f_n(0)=0$ for every $n$; at $x=1$, $f_n(1)=1$ for every $n$. If $0<x<1$, then the sequence $x^n$ decreases to zero. More explicitly, its $N$th tail has supremum $x^N$ and infimum $0$: the terms are at most $x^N$, and $x^k\to0$ as $k\to\infty$. Hence $S_N(x)=x^N$ and $T_N(x)=0$. Taking the infimum of $S_N(x)$ over $N$ and the supremum of $T_N(x)$ over $N$ gives zero. Therefore

\limsup_{n\to\infty}x^n=\liminf_{n\to\infty}x^n= \begin{cases} 0,&0\le x<1,\\ 1,&x=1. \end{cases}

Check: at $x=\frac12$, the tail supremum is $2^{-N}$ and the tail infimum is $0$, so both limiting tail values are $0$. At $x=1$, every tail has supremum and infimum $1$, confirming the separate endpoint value.

When Limsup and Liminf Describe Convergence

For real-valued sequences, finite limsup and liminf provide an exact convergence test. This statement is pointwise: when applied to functions, it applies separately at each $x$, without asserting uniform or almost-everywhere convergence.

Proposition: For a real-valued sequence $(a_n)$, it converges to a finite real number $a$ if and only if $\limsup_n a_n=\liminf_n a_n=a$.

Proof. Suppose first that both tail limits equal the finite number $a$. Let $\varepsilon>0$. Since $\inf_nS_n=a<a+\varepsilon$, some $N_1$ satisfies $S_{N_1}<a+\varepsilon$. Since $\sup_nT_n=a>a-\varepsilon$, some $N_2$ satisfies $T_{N_2}>a-\varepsilon$. Set $N=\max(N_1,N_2)$. The tail suprema decrease and the tail infima increase, so $S_N<a+\varepsilon$ and $T_N>a-\varepsilon$. For every $k\ge N$, $T_N\le a_k\le S_N$. Thus $a-\varepsilon<a_k<a+\varepsilon$, proving $a_n\to a$.

Conversely, suppose $a_n\to a$. Given $\varepsilon>0$, there is an $N$ such that $a-\varepsilon<a_k<a+\varepsilon$ for all $k\ge N$. For every $n\ge N$, this gives $S_n\le a+\varepsilon$ and $T_n\ge a-\varepsilon$. For $n<N$, the larger tail used to define $S_n$ contains the $N$th tail, so $S_n\ge S_N\ge a-\varepsilon$; also $T_n\le T_N\le a+\varepsilon$. It follows that $a-\varepsilon\le\inf_nS_n\le a+\varepsilon$ and $a-\varepsilon\le\sup_nT_n\le a+\varepsilon$. Since this holds for every $\varepsilon>0$, both tail limits equal $a$. $\square$

A Common Pitfall

A common incorrect step is to infer $\sup_k b_k<a$ from $b_k<a$ for every $k$. Take $b_k=1-1/k$ and $a=1$. Each $b_k<1$, but their supremum is $1$, not a number strictly below $1$. The corresponding tail suprema also have infimum $1$, so $\limsup_k b_k=1$, not less than $1$.

The infimum version fails in the same way: $c_k=1/k>0$ for every $k$, but $\inf_kc_k=0$. Thus “all terms are strictly above $a$” does not imply that their infimum is strictly above $a$. To prove a strict inequality, first place a rational threshold strictly between the proposed bound and the tail limit. The measurable level-set formulas use exactly this margin.

Limsup and liminf separate persistent oscillation from convergence: a positive gap between them rules out convergence to a finite value, while equality at a finite value characterizes convergence. Their measurable level sets also make them useful when the terms are measurable functions, even when no pointwise limit has yet been established. These definitions do not involve integration or any convergence in measure or in $L^p$; they describe pointwise tail behavior.

Check Your Understanding

Use the tail definitions and pay attention to strict inequalities in level-set arguments.

For $a_n=(-1)^n$, compute the supremum and infimum of each tail, then find the limsup and liminf.
For $b_n=2+1/n$, find $\limsup b_n$ and $\liminf b_n$ directly from the tail suprema and infima.
Explain why $\{L<a\}$ cannot in general be written as $\bigcup_n\bigcap_{k\ge n}\{f_k<a\}$. Give a numerical sequence demonstrating the failure.
For $f_n(x)=x^n$ on $[0,1]$, identify the points where the limsup and liminf equal $1$.
Prove that if a real sequence has limsup strictly larger than its liminf, it cannot converge to a finite real number.