Pointwise and Almost-Everywhere Limits of Moving Supports

When a Support Moves or Shrinks

An indicator function records membership: it is one on its set and zero elsewhere. For a sequence of such functions, the decisive question at each point is not just whether the sets become small, but how often that point belongs to them. A point that stays in every support gives a different limit from a point that eventually leaves; a point that repeatedly enters and leaves prevents pointwise convergence.

Let $(X,\mathcal{M},\mu)$ be any measure space; no finiteness, sigma-finiteness, or completeness assumption is needed below. Let $E_n\in\mathcal{M}$, and set $f_n=\chi_{E_n}$. These are measurable, real-valued functions taking values in $\{0,1\}$. We use “support” here to mean the set $E_n$ on which the indicator is nonzero, not the topological support of a function.

Definition: The limsup of the sets $E_n$ consists of points that belong to infinitely many of them, and the liminf consists of points that belong to all but finitely many of them: $$ \limsup_{n\to\infty}E_n=\bigcap_{k=1}^{\infty}\bigcup_{n\ge k}E_n,\qquad \liminf_{n\to\infty}E_n=\bigcup_{k=1}^{\infty}\bigcap_{n\ge k}E_n. $$ Both sets are in $\mathcal{M}$, since $\mathcal{M}$ is a sigma-algebra. In particular, $\liminf E_n\subseteq\limsup E_n$.

The limsup and liminf describe two distinct kinds of persistence. Membership in $\liminf E_n$ means the point is eventually always in the supports. A point outside $\limsup E_n$ belongs to only finitely many supports, so it is eventually always outside. The points in between belong to infinitely many supports and also fail to belong to infinitely many; there the indicator sequence alternates between zero and one without settling down.

A Pointwise Convergence Criterion

Theorem: For measurable $E_n\subseteq X$ and $f_n=\chi_{E_n}$, the sequence $(f_n)$ converges pointwise on $X$ if and only if $$ \limsup_{n\to\infty}E_n=\liminf_{n\to\infty}E_n. $$ When it converges, its pointwise limit is $\chi_{\liminf E_n}$. It converges $\mu$-almost everywhere if and only if $\mu(\limsup E_n\setminus\liminf E_n)=0$; in that case $\chi_{\liminf E_n}$ is an almost-everywhere limit.

Proof. Write $L=\liminf E_n$, $U=\limsup E_n$, and $D=U\setminus L$. Fix $x\in X$. If $x\in L$, there is an index $k$ such that $x\in E_n$ for every $n\ge k$. Thus $f_n(x)=1$ eventually and $f_n(x)\to1$. If $x\notin U$, then $x$ belongs to only finitely many $E_n$, so $f_n(x)=0$ eventually and $f_n(x)\to0$. These two cases also show that wherever the sequence converges outside $D$, its limit is $\chi_L(x)$.

Now suppose $x\in D$. Since $x\in U$, it belongs to infinitely many $E_n$, and hence $f_n(x)=1$ infinitely often. Since $x\notin L$, it is not true that $x\in E_n$ for all sufficiently large $n$; therefore $x\notin E_n$ for infinitely many $n$, and $f_n(x)=0$ infinitely often. A real sequence with infinitely many terms equal to zero and infinitely many equal to one cannot converge: convergence to a real number would require both constant subsequences to have that same limit, although their limits are zero and one. Thus precisely the points of $D$ fail to have a pointwise limit. Since $L\subseteq U$, the set $D$ is empty exactly when $L=U$, proving the pointwise statement.

For almost-everywhere convergence, if $\mu(D)=0$, the preceding pointwise analysis gives convergence to the measurable function $\chi_L$ at every point of $X\setminus D$. Conversely, if $f_n$ converges almost everywhere, there is a measurable null set $N$ outside which it converges. No point of $D$ can be outside $N$, because convergence fails at each such point. Hence $D\subseteq N$, and monotonicity of measure gives $\mu(D)=0$. This proves both directions without requiring the measure space to be complete. $\square$

Corollary (Nested supports): If $E_{n+1}\subseteq E_n$ for every $n$, then $\chi_{E_n}$ converges pointwise everywhere to $\chi_{\cap_{n=1}^{\infty}E_n}$. If that intersection is null, the limit is zero almost everywhere.

Proof. Fix $x\in X$. If $x\in\bigcap_n E_n$, then $f_n(x)=1$ for every $n$. If $x\notin\bigcap_n E_n$, then there is an index $k$ with $x\notin E_k$. The nesting $E_n\subseteq E_k$ for all $n\ge k$ implies $x\notin E_n$ for every $n\ge k$, so $f_n(x)=0$ eventually. This gives the stated limit at every point. If the intersection is null, the limit is zero outside a null set, which is exactly almost-everywhere convergence to zero. $\square$

Worked Interval Examples

Worked Example: Nested Intervals That Retain One Point

On $(\mathbb{R},\mathcal{L},m)$, let $E_n=[0,1/n]$ and $f_n=\chi_{E_n}$. Find the pointwise limit and decide whether $f_n\to0$ almost everywhere. The intervals include both endpoints: a point $x$ belongs to $E_n$ precisely when $0\le x\le1/n$.

At $x=0$, the membership condition holds for every $n$, so $f_n(0)=1$ for all $n$. If $x<0$, it fails for every $n$, so $f_n(x)=0$ for every $n$. If $x>0$, choose an integer $N>1/x$. For every $n\ge N$, $1/n\le1/N<x$, so $x\notin[0,1/n]$ and $f_n(x)=0$. Consequently, the pointwise limit is $\chi_{\{0\}}$, everywhere on $\mathbb{R}$.

The sets are nested and their intersection is $\{0\}$. Also, $m(\{0\})=0$, while $m(E_n)=1/n$. Thus the pointwise limit is zero almost everywhere, even though the sequence does not converge to zero at $x=0$. Check: substituting $x=0$ gives $f_n(0)=1$ for every $n$, while substituting $x=1$ gives $f_1(1)=1$ and $f_n(1)=0$ for every $n\ge2$. These values agree with the stated limit at both points.

Worked Example: Shrinking Supports That Alternate Between Two Points

On $(\mathbb{R},\mathcal{L},m)$, define $$ E_n= \begin{cases} [0,1/n],&n\text{ even},\\ [2,2+1/n],&n\text{ odd}, \end{cases} \qquad f_n=\chi_{E_n}. $$ Determine where $f_n$ converges. The closed endpoints are included in each membership test.

The point $0$ belongs to every even-indexed interval and to no odd-indexed interval. Thus $f_n(0)=1$ for even $n$ and $f_n(0)=0$ for odd $n$, so the sequence does not converge at $0$. Similarly, $2$ belongs to every odd-indexed interval and no even-indexed interval; $f_n(2)$ equals one for odd $n$ and zero for even $n$. Hence convergence also fails at $2$.

Fix any $x\notin\{0,2\}$. The distances $d_0=|x|$ and $d_2=|x-2|$ are both positive. Choose $N$ so large that $1/N<\min(d_0,d_2)$. For $n\ge N$, the interval of length $1/n$ starting at its corresponding center, either 0 or 2, cannot contain $x$: its distance from that center is at least $\min(d_0,d_2)>1/N\ge1/n$. Therefore $f_n(x)=0$ for all $n\ge N$. The sequence converges to zero at every point except $0$ and $2$; since this exceptional set has Lebesgue measure zero, convergence to zero holds almost everywhere.

For each $n$, $m(E_n)=1/n$, since each support is a closed interval of length $1/n$. The common-threshold membership pattern is also consistent with the criterion: $0$ and $2$ each lie in infinitely many supports but not eventually all of them, while every other point lies in only finitely many. Check: $m(E_2)=1/2$, $m(E_3)=1/3$, and both are obtained by subtracting the interval endpoints; at the exceptional points the even and odd subsequences have distinct limits, 1 and 0.

Worked Example: Zero-Measured Supports and a Pointwise Exception

On $(\mathbb{R},\mathcal{L},m)$, let $E_n=\{0\}$ for odd $n$, and $E_n=\varnothing$ for even $n$. Each set is measurable and has measure zero. At $x=0$, $f_n(0)$ alternates between one and zero, so there is no pointwise limit there. At every $x\ne0$, $f_n(x)=0$ for all $n$. Thus the sequence converges to zero almost everywhere but not pointwise everywhere.

Here $m(E_n)=0$ for every $n$, because both the empty set and a singleton have Lebesgue measure zero. Nevertheless, $0\in\limsup E_n$ because it occurs for infinitely many odd indices, and $0\notin\liminf E_n$ because it is absent at every even index. Therefore $\limsup E_n\setminus\liminf E_n=\{0\}$, exactly the exceptional set predicted by the theorem. Check: $f_1(0)=1$, $f_2(0)=0$, and $f_3(0)=1$, so the sequence at zero cannot converge; for $x=1$, all three values and indeed every $f_n(1)$ equal zero.

A Common Misreading

A shrinking measure is not itself a pointwise convergence argument. For instance, in the first interval example $m(E_n)=1/n\to0$, but $f_n(0)=1$ for every $n$. The correct pointwise test is membership at the specific point: here $0$ belongs to every support, so the limit at $0$ is one. The measure calculation supports an almost-everywhere conclusion only after identifying the exceptional set $\{0\}$ and checking that its measure is zero.

A related error is to confuse “infinitely often” with “eventually always.” In the alternating-interval example, $0$ lies in infinitely many supports, but it does not lie in all sufficiently late supports. It belongs to the limsup but not the liminf, and the indicator oscillates there. By contrast, membership in the liminf guarantees eventual value one. These distinctions apply to any measurable sequence of supports, not only intervals. They do not require finite measure, and the pointwise criterion does not rely on any interchange of limits with integrals.

Check Your Understanding

Use the limsup and liminf membership descriptions to justify each answer.

On $\mathbb{R}$, let $E_n=[-1/n,1/n]$. Find the pointwise limit of $\chi_{E_n}$, and state whether the limit is zero almost everywhere.
Suppose $E_n$ is increasing, meaning $E_n\subseteq E_{n+1}$ for every $n$. At which points does $\chi_{E_n}$ converge to one?
For $E_n=[3,3+1/n]$ when $n$ is even and $E_n=[5,5+1/n]$ when $n$ is odd, identify every point where pointwise convergence fails and give the almost-everywhere limit.
State the exact set whose measure must be zero for $\chi_{E_n}$ to converge almost everywhere.
Give a reason that $m(E_n)\to0$ alone does not imply pointwise convergence at every point.

Pointwise and Almost-Everywhere Limits of Moving Supports

What You'll Learn