Countable Suprema and Infima of Measurable Functions

Why Countable Suprema and Infima Are Measurable

Pointwise operations on finitely many measurable functions preserve measurability, but sequences introduce a different question: do the pointwise supremum and infimum remain measurable? The decisive fact is that their strict level sets can be written using countable unions or intersections of the level sets of the original functions. Countability matters because a sigma-algebra is closed under countable unions and intersections.

Unless an example specifies otherwise, let $(X,\mathcal{M},\mu)$ be any measure space; no assumption of finiteness, sigma-finiteness, or completeness is needed. Let $f_n:X\to\mathbb{R}$ be real-valued measurable functions, with measurability understood through measurable preimages of Borel sets. In particular, the sets $\{x:f_n(x)>a\}$ and $\{x:f_n(x)<a\}$ belong to $\mathcal{M}$ for every real $a$. We use the half-line criterion: a real-valued function $h$ is measurable if $\{x:h(x)>a\}\in\mathcal{M}$ for every real $a$; equivalently, one can test $\{x:h(x)<a\}$. These are pointwise statements about functions, not assertions about integrability or convergence in measure.

Definition: The pointwise supremum and infimum of the sequence are $S(x)=\sup_{n\ge1} f_n(x)$ and $I(x)=\inf_{n\ge1} f_n(x)$. They are initially allowed to take the values $+\infty$ or $-\infty$. To call $S$ or $I$ a real-valued measurable function, its value must be finite at every $x\in X$.

Strict Level Sets

The level-set identities below are the main tool. They distinguish carefully between a strict inequality for a supremum or infimum and a non-strict inequality for an individual term.

Proposition (Strict-level-set identities): For real-valued functions $f_n$ on $X$, with $S(x)=\sup_n f_n(x)$ and $I(x)=\inf_n f_n(x)$, the following identities hold for each real $a$: $$ \{S>a\}=\bigcup_{n=1}^{\infty}\{f_n>a\},\qquad \{I<a\}=\bigcup_{n=1}^{\infty}\{f_n<a\}. $$ If $S$ and $I$ are finite everywhere, then also $$ \{S<a\}=\bigcup_{\substack{q\in\mathbb{Q}\\q<a}}\ \bigcap_{n=1}^{\infty}\{f_n\le q\},\qquad \{I>a\}=\bigcup_{\substack{q\in\mathbb{Q}\\q>a}}\ \bigcap_{n=1}^{\infty}\{f_n\ge q\}. $$

Proof. Fix $x\in X$. The inequality $S(x)>a$ holds exactly when at least one term $f_n(x)$ is greater than $a$: if every $f_n(x)\le a$, then $a$ is an upper bound and $S(x)\le a$; if some $f_n(x)>a$, then the supremum is at least that term and exceeds $a$. This proves the first identity. The infimum identity follows by the corresponding argument: $I(x)<a$ holds exactly when some $f_n(x)<a$, since if every term is at least $a$, then $I(x)\ge a$.

For the third identity, suppose $S(x)<a$. By density of the rationals, there is a rational $q$ with $S(x)<q<a$. Since $f_n(x)\le S(x)<q$ for every $n$, the point $x$ belongs to $\bigcap_n\{f_n\le q\}$, and hence to the displayed union. Conversely, if $x\in\bigcap_n\{f_n\le q\}$ for some rational $q<a$, then $q$ is an upper bound for all $f_n(x)$, so $S(x)\le q<a$. For the fourth identity, if $I(x)>a$, choose a rational $q$ with $a<q<I(x)$. Then $f_n(x)\ge I(x)>q$ for every $n$. Conversely, if $f_n(x)\ge q$ for every $n$ and some $q>a$, then $I(x)\ge q>a$. The assumed finiteness ensures the strict inequalities used to choose these rational thresholds apply to real values. $\square$

Each union or intersection here is countable: the indices $n$ are natural numbers, and the rationals are countable. Because each $f_n$ is measurable, its strict and non-strict level sets are in $\mathcal{M}$. The sigma-algebra properties therefore make all four sets measurable whenever the corresponding formulas apply.

Measurability Theorem

Theorem: Let $(X,\mathcal{M},\mu)$ be any measure space and let $(f_n)$ be real-valued measurable functions on $X$. If $S(x)=\sup_n f_n(x)$ is finite for every $x$, then $S:X\to\mathbb{R}$ is measurable. If $I(x)=\inf_n f_n(x)$ is finite for every $x$, then $I:X\to\mathbb{R}$ is measurable. The two finiteness assumptions are separate.

Proof. Assume first that $S$ is finite everywhere. For every real $a$, the first strict-level-set identity gives $\{S>a\}=\bigcup_n\{f_n>a\}$. Each set in the union belongs to $\mathcal{M}$, and a sigma-algebra is closed under countable unions. Thus $\{S>a\}\in\mathcal{M}$ for every $a$, so the half-line criterion proves that the real-valued function $S$ is measurable.

Now assume that $I$ is finite everywhere. For every real $a$, the second identity gives $\{I<a\}=\bigcup_n\{f_n<a\}$. Each term is measurable, so this countable union belongs to $\mathcal{M}$. The equivalent lower-half-line criterion proves that $I$ is measurable. Neither argument uses a measure or an integral; it uses only the sigma-algebra and the level-set criterion. $\square$

The finiteness conditions cannot simply be omitted if the conclusion is stated for real-valued functions. For example, on a nonempty space the constant sequence $f_n(x)=n$ consists of measurable real-valued functions, but its pointwise supremum is $+\infty$ at every point, not a real-valued function. Its strict upper level sets still satisfy the union identity; treating extended-real-valued functions fully is a separate matter.

Worked Examples

Worked Example: A Supremum That Is Not Attained

On $X=[0,1]$ with its Lebesgue sigma-algebra and Lebesgue measure $m$, let $f_n(x)=x-\frac1n$, for $n\ge1$. Each $f_n$ is continuous and therefore measurable. Find the pointwise supremum and infimum and check their level sets.

For fixed $x$, $1/n$ decreases to $0$, so $x-1/n$ increases to $x$. Every term is strictly less than $x$, while the terms approach $x$; hence $S(x)=x$, and this supremum is not attained by any finite $n$. The smallest term occurs at $n=1$, giving $I(x)=x-1$. Both functions are finite on $[0,1]$.

For any real $a$, the supremum identity gives $\{S>a\}=\bigcup_n\{x:x-1/n>a\}$. If $x>a$, choose an integer $n>1/(x-a)$; then $x-1/n>a$. If $x\le a$, no $n$ can satisfy that inequality. Thus the union is $\{x\in[0,1]:x>a\}$, the strict upper level set of $S(x)=x$. For the infimum, $\{I<a\}=\bigcup_n\{x:x-1/n<a\}$. This union equals $\{x\in[0,1]:x-1<a\}$: the $n=1$ term already gives that set, and every other term is at least $x-1$.

Check: at $x=0$, $f_n(0)=-1/n$, so the supremum is $0$ and the infimum is $-1$. At $x=1$, the terms are $1-1/n$, so the supremum is $1$ and the infimum is $0$. These agree with $S(x)=x$ and $I(x)=x-1$.

Worked Example: Suprema and Infima of Nested Indicators

On $(\mathbb{R},\mathcal{L},m)$, define $E_n=(-1/n,\,2+1/n)$ and $f_n=\chi_{E_n}$. Determine the pointwise supremum and infimum, paying attention to open endpoints.

Since $1/(n+1)<1/n$, the left endpoint $-1/(n+1)$ is greater than $-1/n$, and the right endpoint $2+1/(n+1)$ is less than $2+1/n$. Therefore $E_{n+1}\subset E_n$. The union is $E_1=(-1,3)$: every later interval lies in $E_1$, and $E_1$ itself is included in the union. The intersection is $[0,2]$. Indeed, if $0\le x\le2$, then $-1/n<x<2+1/n$ for every $n$. If $x<0$, choose $n$ so large that $1/n\le -x$; then $-1/n<x$ fails. If $x>2$, choose $n$ so large that $1/n\le x-2$; then $x<2+1/n$ fails.

For indicator functions, the supremum is $1$ exactly on the union of the sets, while the infimum is $1$ exactly on their intersection. Consequently, $S=\chi_{(-1,3)}$ and $I=\chi_{[0,2]}$. These functions are measurable because the indicated sets belong to $\mathcal{L}$.

Check: at $x=3/2$, the point lies in every $E_n$, so all $f_n(3/2)=1$, and both $S(3/2)$ and $I(3/2)$ equal $1$. At $x=5/2$, the point lies in $E_1=(-1,3)$, but not in the intersection $[0,2]$; hence the supremum is $1$ and the infimum is $0$. At $x=3$, no $E_n$ contains $x$, because each right endpoint is strictly less than $3$, so both values are $0$.

Worked Example: A Sequence Increasing to a Discontinuous Supremum

On $X=[0,1]$ with Lebesgue measure $m$, set $f_n(x)=\min(nx,1)$. Each function is continuous, hence measurable. Compute its pointwise supremum and infimum.

For $x\in[0,1]$, increasing $n$ increases $nx$, so $\min(nx,1)$ is nondecreasing in $n$. At $x=0$, every term is $0$. If $x>0$, choose an integer $n\ge1/x$; then $nx\ge1$, so $f_n(x)=1$. Since no term exceeds $1$, the supremum is $S(0)=0$ and $S(x)=1$ for $0<x\le1$. The infimum of this increasing sequence is its first term, $I(x)=f_1(x)=x$.

For example, at the strict threshold $a=1/2$, $\{S>a\}=(0,1]$. Indeed, $x=0$ is in none of the sets $\{f_n>1/2\}$; for each $x>0$, choose $n>1/(2x)$, which gives $nx>1/2$ and hence $f_n(x)>1/2$. Thus the countable union $\bigcup_n\{f_n>1/2\}$ is exactly $(0,1]$. For the infimum, $\{I<1/2\}=[0,1/2)$, directly from $I(x)=x$.

Check: at $x=0$, both the supremum and infimum are $0$. At $x=1/4$, $f_1(1/4)=1/4$, $f_2(1/4)=1/2$, and $f_4(1/4)=1$; thus the infimum is $1/4$ and the supremum is $1$, as the formulas predict. At $x=1$, every term is $1$, so both values are $1$.

A Common Mistake and Why the Result Matters

A frequent error is to use a union when testing whether an infimum is above a threshold. The correct identity for $\{I>a\}$, when the infimum is finite, is the rational-threshold formula involving intersections proved above. For a simpler non-strict test, $\{I\ge a\}=\bigcap_n\{f_n\ge a\}$: every term must be at least $a$.

For a concrete incorrect calculation, take $X=\{x_0\}$ with any sigma-algebra and let $f_1(x_0)=0$, $f_n(x_0)=2$ for every $n\ge2$. The infimum is $I(x_0)=0$, so $\{I>1\}$ is empty. The wrong union calculation $\bigcup_n\{f_n>1\}$ is not empty, because $x_0\in\{f_2>1\}$. The correction is that an infimum exceeds $1$ only if every term exceeds $1$; here $\bigcap_n\{f_n>1\}$ is empty. The corresponding union identity applies instead to $\{I<1\}$.

These results are useful whenever a function is defined pointwise by optimizing over a sequence of measurable functions. They also explain why the proof is specific to a countable family: the required unions and intersections are guaranteed to stay in a sigma-algebra. The theorem establishes measurability only; it does not assert boundedness, integrability, or any convergence of integrals.

Check Your Understanding

In each question, distinguish strict from non-strict inequalities and state any finiteness assumption needed to call the resulting function real-valued.

Prove directly that $\{\sup_n f_n>a\}=\bigcup_n\{f_n>a\}$ for real-valued functions $f_n$.
On $[0,1]$, let $f_n(x)=x+1/n$. Find the pointwise supremum and infimum, and state which is attained.
For $E_n=[0,1/n)$ in $\mathbb{R}$, find the supremum and infimum of $\chi_{E_n}$, including the value at $x=0$.
Give a sequence of measurable real-valued functions whose pointwise supremum is not finite everywhere, and explain why the real-valued measurability theorem does not apply as stated.
Why does the identity $\{I<a\}=\bigcup_n\{f_n<a\}$ use a union rather than an intersection?