Limits of Measurable Functions

Why Limits Matter for Measurability

The preceding tutorial showed how nonnegative measurable functions can be approximated pointwise by simple measurable functions. More generally, pointwise limits are a basic way to build new functions from sequences of known ones. The key question is whether measurability survives this passage to a limit. It does, and the proof reveals why countable operations are central to measure theory.

We work on a measurable space $(X,\mathcal{F})$. For a real-valued function $f$, measurability can be tested through its threshold sets: in particular, $\{x\in X:f(x)>a\}$ is measurable for every real $a$ if and only if $f$ is measurable, by the Threshold Characterization of Measurability. We will use this characterization to turn a question about function limits into a question about countable unions and intersections of measurable sets.

Pointwise Limits of Measurable Functions

Suppose a sequence of measurable functions $f_n:X\to\mathbb{R}$ converges at every point to a finite real-valued function $f$. To check that $f$ is measurable, fix a threshold $a$. If $f(x)>a$, there is a rational number $q$ strictly between $a$ and $f(x)$. Convergence then forces all sufficiently late values $f_n(x)$ to exceed $q$. Conversely, if all sufficiently late values exceed some $q>a$, their limit cannot be below $q$, and therefore must exceed $a$.

This reasoning gives an exact description of the threshold set:

\{x:f(x)>a\} = \bigcup_{\substack{q\in\mathbb{Q}\\q>a}} \bigcup_{N=1}^{\infty} \bigcap_{n=N}^{\infty}\{x:f_n(x)>q\}.

Both unions and the intersection here are countable. That is what allows the right-hand side to remain measurable.

Theorem (Pointwise Limits of Measurable Functions): Let $(X,\mathcal{F})$ be a measurable space. Suppose $f_n:X\to\mathbb{R}$ is measurable for every positive integer $n$, and $f_n(x)\to f(x)\in\mathbb{R}$ for every $x\in X$. Then $f:X\to\mathbb{R}$ is measurable.

Proof. Fix $a\in\mathbb{R}$. For each rational $q>a$ and each positive integer $N$, define

E_{q,N}=\bigcap_{n=N}^{\infty}\{x\in X:f_n(x)>q\}.

Every set $\{f_n>q\}$ is measurable because $f_n$ is measurable. Since $\mathcal{F}$ is closed under countable intersections, $E_{q,N}\in\mathcal{F}$. Closure under countable unions then shows that the set

E=\bigcup_{\substack{q\in\mathbb{Q}\\q>a}}\bigcup_{N=1}^{\infty}E_{q,N}

is measurable. We verify that $E=\{f>a\}$. If $f(x)>a$, choose a rational $q$ with $a<q<f(x)$. Since $f_n(x)\to f(x)$, take $\varepsilon=(f(x)-q)/2>0$. There is an $N$ such that for every $n\geq N$,

f_n(x)>f(x)-\varepsilon=\frac{f(x)+q}{2}>q.

Thus $x\in E_{q,N}\subseteq E$. In the other direction, if $x\in E$, then for some rational $q>a$ and some $N$, $f_n(x)>q$ for every $n\geq N$. Taking the limit gives $f(x)\geq q>a$, so $x\in\{f>a\}$. Therefore $E=\{f>a\}$, which is measurable. This holds for every real $a$; the Threshold Characterization of Measurability now gives that $f$ is measurable. $\square$

Worked Example: A Pointwise Limit That Is Discontinuous

For each positive integer $n$, define $g_n:\mathbb{R}\to\mathbb{R}$ by

g_n(x)=\max(1-n|x|,0).

Each $g_n$ is continuous, so it is Borel measurable. At $x=0$, $g_n(0)=\max(1,0)=1$ for every $n$. If $x\neq 0$, choose $N$ so large that $N\geq 1/|x|$. For every $n\geq N$, $1-n|x|\leq 0$, so $g_n(x)=0$. Consequently, the pointwise limit is

g(x)= \begin{cases} 1,&x=0,\\ 0,&x\neq 0. \end{cases} =\mathbf{1}_{\{0\}}(x).

This limit is discontinuous at $0$, but it is measurable: $\{0\}$ is a Borel set, and the Measurability of an Indicator Function Theorem applies. The example shows that pointwise limits preserve measurability, not continuity.

Limsup and Liminf of Measurable Functions

A sequence of functions need not converge everywhere. Even then, two extended real-valued functions summarize its eventual upper and lower behavior. For each $x$, define the tail supremum and tail infimum, and then take their limits as the tails begin farther out.

Definition (Pointwise Limsup and Liminf): For a sequence of real-valued functions $(f_n)$, define $$ \limsup_{n\to\infty}f_n(x)=\inf_{N\geq 1}\sup_{n\geq N}f_n(x), \qquad \liminf_{n\to\infty}f_n(x)=\sup_{N\geq 1}\inf_{n\geq N}f_n(x). $$ These values are allowed to be $+\infty$ or $-\infty$. An extended real-valued function is called measurable when its strict upper and lower threshold sets are measurable for every real threshold.

The limsup records the values that can remain as large as a given threshold arbitrarily far along the sequence. The liminf records the corresponding lower behavior. If a sequence converges to a finite value at $x$, both quantities equal that limit at $x$. Their measurability does not require convergence.

Theorem (Measurability of Limsup and Liminf): If $f_n:X\to\mathbb{R}$ is measurable for every $n$, then $\limsup_{n\to\infty}f_n$ and $\liminf_{n\to\infty}f_n$ are measurable extended real-valued functions.

Proof. Fix $a\in\mathbb{R}$. For the limsup, we claim

\left\{x:\limsup_{n\to\infty}f_n(x)>a\right\} = \bigcup_{\substack{q\in\mathbb{Q}\\q>a}} \bigcap_{N=1}^{\infty} \bigcup_{n=N}^{\infty}\{x:f_n(x)>q\}.

If $\limsup_n f_n(x)>a$, choose a rational $q$ strictly between $a$ and $\limsup_n f_n(x)$. For every $N$, the supremum of the tail $\{f_n(x):n\geq N\}$ is greater than $q$. Hence there is some $n\geq N$ with $f_n(x)>q$, so $x$ belongs to the right-hand side. Conversely, if $x$ belongs to the right-hand side, then for some $q>a$, every tail contains a value greater than $q$. Thus every tail supremum is at least $q$, and their infimum is at least $q>a$. This proves the claimed identity.

Every set $\{f_n>q\}$ on the right is measurable. The displayed expression uses only countable unions and intersections, so the limsup’s strict upper threshold set is measurable.

For the liminf, the corresponding identity is

\left\{x:\liminf_{n\to\infty}f_n(x)<a\right\} = \bigcup_{\substack{q\in\mathbb{Q}\\q<a}} \bigcap_{N=1}^{\infty} \bigcup_{n=N}^{\infty}\{x:f_n(x)<q\}.

If the liminf is less than $a$, choose a rational $q$ strictly between it and $a$. Every tail has infimum less than $q$, so each tail contains a value less than $q$. Conversely, if every tail contains a value less than some $q<a$, every tail infimum is at most $q$, and the supremum of those tail infima is at most $q<a$. This proves the identity. Its right-hand side is measurable by countable closure, so the liminf’s strict lower threshold sets are measurable. Together these two conclusions give measurability of both extended real-valued functions: indeed, $\{x:\limsup_{n\to\infty}f_n(x)<a\}=X\setminus\bigcap_{k=1}^{\infty}\{x:\limsup_{n\to\infty}f_n(x)>a-1/k\}$ and $\{x:\liminf_{n\to\infty}f_n(x)>a\}=X\setminus\bigcap_{k=1}^{\infty}\{x:\liminf_{n\to\infty}f_n(x)<a+1/k\}$. Each is the complement of a countable intersection of measurable threshold sets. $\square$

Worked Example: A Sequence with Different Limsup and Liminf

Define $h_n:\mathbb{R}\to\mathbb{R}$ by $h_n(x)=(-1)^n+x/n$. Each $h_n$ is continuous. Fix $x$. Along even indices $n=2k$,

h_{2k}(x)=1+\frac{x}{2k}\longrightarrow 1,

and along odd indices $n=2k+1$,

h_{2k+1}(x)=-1+\frac{x}{2k+1}\longrightarrow -1.

The even and odd subsequences therefore show that $\limsup_n h_n(x)=1$ and $\liminf_n h_n(x)=-1$ for every $x$. The sequence does not converge at any point, but its limsup and liminf are constant measurable functions, as the theorem guarantees.

Limits Built from Indicators

Pointwise limits also make it possible to describe measurable sets through sequences of simpler measurable sets. For instance, enumerate the rational numbers as $q_1,q_2,\ldots$, and set $A_n=\{q_1,\ldots,q_n\}$. Each $A_n$ is finite and hence Borel measurable. The indicators $\mathbf{1}_{A_n}$ are measurable, and their pointwise limit is the indicator of $\mathbb{Q}$: every rational eventually belongs to all later $A_n$, while no irrational belongs to any $A_n$.

Worked Example: Approximating the Indicator of the Rationals

Let $r_n=\mathbf{1}_{A_n}$, where $A_n=\{q_1,\ldots,q_n\}$ for an enumeration of $\mathbb{Q}$. If $x=q_j$, then $r_n(x)=1$ for every $n\geq j$, so $r_n(x)\to 1$. If $x\notin\mathbb{Q}$, then $r_n(x)=0$ for every $n$, so $r_n(x)\to 0$. Therefore

\lim_{n\to\infty}r_n(x)=\mathbf{1}_{\mathbb{Q}}(x).

The set $\mathbb{Q}$ is countable and Borel measurable, so this limit is measurable. This construction is a direct application of the Pointwise Limits of Measurable Functions Theorem; it also shows how countable descriptions of sets can arise from limits of indicators.

What Pointwise Convergence Does Not Say

The word “pointwise” matters. The stage at which $f_n(x)$ becomes close to its limit may depend on $x$. In the spike example, the functions $g_n$ converge pointwise to $\mathbf{1}_{\{0\}}$, but the convergence is not uniform: for every $n$, $g_n(1/(2n))=1/2$, while the limit at $1/(2n)$ is $0$. Thus the error is at least $1/2$ somewhere for every $n$.

Nor does pointwise convergence alone imply convergence of measures, integrals, or other quantities associated with the functions. Such conclusions require additional hypotheses and separate convergence results. The result established here is specifically about the measurability of the limit function. In the opposite direction, the Increasing Simple Approximation Theorem from the preceding tutorial supplies sequences of measurable simple functions converging pointwise to nonnegative measurable functions. The theorem in this tutorial explains why pointwise limits are a reliable way to construct measurable functions, while not claiming that every sequence of measurable functions converges.

Key takeaway: A finite pointwise limit of measurable real-valued functions is measurable because its threshold sets can be expressed using countable unions and intersections. Even without convergence, the pointwise limsup and liminf of measurable functions are measurable extended real-valued functions.

Check Your Understanding

Use the threshold-set arguments and examples to answer the following questions.

Why can a rational number between $a$ and $f(x)$ be used when $f(x)>a$ in the proof of the pointwise limit theorem?
Which countable set operations appear in the expression for $\{f>a\}$ when $f_n\to f$?
For $h_n(x)=(-1)^n+x/n$, what are the pointwise limsup and liminf, and why does the sequence fail to converge?
Why is the pointwise limit of the continuous spike functions discontinuous at $0$ but still measurable?
What additional conclusion about integrals or measures follows from pointwise convergence alone?

What You'll Learn