When the Supremum Is Not Attained

What It Means for a Supremum to Be Attained

For a nonnegative measurable function, the Lebesgue integral is defined as a supremum over simple functions below it. A supremum need not be one of the values being considered. This distinction matters even when the integral is finite: simple minorants can have integrals arbitrarily close to the supremum, while every individual minorant still falls strictly short.

Throughout, $(X,\mathcal{M},\mu)$ is a measure space, not assumed finite, complete, or $\sigma$-finite. Functions under discussion are measurable and take values in $[0,\infty]$. A simple function here means a measurable function taking finitely many finite values in $[0,\infty)$. A simple minorant of $f$ is such a function $\varphi$ satisfying $0\le\varphi\le f$ everywhere. The integral of a simple function is defined by summing its values times the measures of their disjoint level sets.

Definition: The nonnegative integral of $f$ is the extended-real supremum $$ \int f\,d\mu=\sup\left\{\int\varphi\,d\mu:\ \varphi\text{ is simple and }0\le\varphi\le f\right\}. $$ The supremum is attained if at least one simple minorant $\varphi$ has $\int\varphi\,d\mu=\int f\,d\mu$. Otherwise it is not attained. The supremum may be finite or $+\infty$.

We will use the level-set formula for a nonnegative simple function. If $\varphi$ has distinct positive values $c_1<\cdots<c_k$, then $\int\varphi\,d\mu=\int_0^\infty\mu(\{\varphi>t\})\,dt$, where the integral in $t$ is just the area under a finite step function. To verify the identity, on each interval between consecutive values the set $\{\varphi>t\}$ is constant; summing the interval lengths times those set measures telescopes to $\sum_{j=1}^k c_j\mu(\{\varphi=c_j\})$, the defining simple-function integral. A zero value contributes nothing.

A Non-Simple Function on an Interval

Worked Example: The Function f(x) = x on [0,1]

Take $X=[0,1]$, $\mathcal{M}$ to be the Lebesgue $\sigma$-algebra on this interval, and $\mu$ to be Lebesgue measure $m$. Let $f(x)=x$. We compute its defining supremum and show that no simple minorant attains it.

First let $\varphi$ be any simple minorant. Because $\varphi\le x\le1$, its values lie in $[0,1]$. For $0\le t\le1$, the pointwise inequality implies $\{\varphi>t\}\subseteq\{x>t\}$, and $m(\{x>t\})=1-t$. For $t>1$, both level sets are empty. The level-set formula therefore gives \[ \int\varphi\,dm=\int_0^1m(\{\varphi>t\})\,dt\le\int_0^1(1-t)\,dt=\frac12. \] This proves that the supremum is at most $1/2$.

To prove the inequality is strict for each individual $\varphi$, note that the set where $\varphi=1$ is contained in $\{x=1\}$, which has measure zero. Among the finitely many values of $\varphi$ that are less than $1$, let $a$ be the largest, including $0$ in this finite collection if necessary. Then $a<1$, and $\varphi\le a$ almost everywhere. Consequently $m(\{\varphi>t\})=0$ for $a<t<1$. The level-set bound on $[0,a]$ and this vanishing on $(a,1)$ imply \[ \int\varphi\,dm\le\int_0^a(1-t)\,dt =\frac12-\frac{(1-a)^2}{2}<\frac12. \] Thus no simple minorant attains $1/2$.

It remains to show that $1/2$ really is the supremum. For each integer $n\ge1$, define $\varphi_n(x)=2^{-n}\lfloor2^nx\rfloor$. This is simple, is nonnegative, and satisfies $\varphi_n(x)\le x$. On each interval $[k/2^n,(k+1)/2^n)$, for $k=0,\ldots,2^n-1$, it equals $k/2^n$; at $x=1$ it equals $1$, on a singleton of measure zero. Hence \[ \int\varphi_n\,dm =\sum_{k=0}^{2^n-1}\frac{k}{2^n}\frac1{2^n} =\frac{1}{4^n}\frac{(2^n-1)2^n}{2} =\frac{2^n-1}{2^{n+1}}. \] As $n\to\infty$, this tends to $1/2$. Since all simple minorants have integral at most $1/2$, the supremum is exactly $1/2$, and the strict inequality above proves it is not attained.

Check: At $n=1$, the formula gives $(2-1)/4=1/4$. At $n=2$, it gives $(4-1)/8=3/8$; directly the three nonzero interval contributions are $(1/4)(1/4)+(2/4)(1/4)+(3/4)(1/4)=6/16=3/8$. The values approach $1/2$ from below, in agreement with the proved bound.

Why a Positive Gap Gives a Better Minorant

The interval example illustrates a useful way to rule out attainment: if a proposed minorant leaves a positive gap on a set of positive measure, it can be raised there. The finiteness condition in the next proposition is essential to conclude that the integral strictly increases.

Proposition: Suppose $\varphi$ is a simple minorant of a measurable $f:X\to[0,\infty]$. If there are $\varepsilon>0$ and $E\in\mathcal{M}$ with $0<\mu(E)<\infty$, $f\ge\varphi+\varepsilon$ on $E$, and $\int\varphi\,d\mu<\infty$, then $\psi=\varphi+\varepsilon\chi_E$ is a simple minorant and $\int\psi\,d\mu>\int\varphi\,d\mu$.

Proof. The sum $\psi$ takes only finitely many finite values, is measurable, and is nonnegative. On $E$, the assumed gap gives $\psi\le f$; outside $E$, $\psi=\varphi\le f$. Thus $\psi$ is a simple minorant. Refine the finite level-set partitions of $\varphi$ and $\chi_E$ to their common disjoint partition. On each piece, the value of $\psi$ is the value of $\varphi$ plus $\varepsilon$ if the piece lies in $E$, and is otherwise unchanged. Summing values times measures yields $\int\psi\,d\mu=\int\varphi\,d\mu+\varepsilon\mu(E)$. The first term is finite by hypothesis, while $\varepsilon\mu(E)$ is finite and strictly positive. Therefore the sum is strictly greater than $\int\varphi\,d\mu$. $\square$

For $f(x)=x$, any proposed simple minorant has a positive gap except possibly where $x$ equals one of its finitely many values. Those exceptional points form a finite set and have measure zero. The positive-gap set is therefore of measure one. It is the countable union of measurable sets $\{x:f(x)-\varphi(x)\ge1/j\}$, $j\ge1$. At least one such set must have positive measure: if all had measure zero, their countable union would have measure zero. It also has finite measure because it lies in $[0,1]$. Every simple minorant on this interval has finite integral, since it is bounded and the interval has measure one. The proposition then shows that each can be improved. This independently confirms why no minorant can be maximal here.

Attainment and Nonattainment in Contrasting Cases

Worked Example: A Simple Function Attains the Supremum

On $[0,1]$ with Lebesgue measure, take $f=\chi_{[0,1/2]}$. It is itself simple and is a minorant of $f$, with integral $m([0,1/2])=1/2$. For any simple $\varphi\le f$, the sets $\{\varphi>t\}$, for $0\le t<1$, are contained in $[0,1/2]$; for $t\ge1$ they are empty. The level-set formula gives \[ \int\varphi\,dm=\int_0^1m(\{\varphi>t\})\,dt \le\int_0^1\frac12\,dt=\frac12. \] Thus the supremum is $1/2$, and it is attained by $f$ itself.

Check: Substituting $\varphi=f$ gives $\int f\,dm=1\cdot m([0,1/2])=1/2$, equal to the upper bound just proved.

Worked Example: An Infinite Supremum Never Reached

Let $X=\mathbb{N}=\{1,2,\ldots\}$, let $\mathcal{M}$ be all subsets of $\mathbb{N}$, and let $\mu$ be counting measure. Set $f(n)=1/n$. For $N\ge1$, define $\varphi_N(n)=1/n$ when $1\le n\le N$, and $\varphi_N(n)=0$ when $n>N$. This is a simple minorant: it has finitely many values and finite support. Its integral is \[ \int\varphi_N\,d\mu=\sum_{n=1}^N\frac1n. \] These partial sums are unbounded. Indeed, for each $j\ge1$, the block $2^{j-1}<n\le2^j$ contains $2^{j-1}$ terms, each at least $1/2^j$, so its sum is at least $2^{j-1}/2^j=1/2$. Including the first term gives $\sum_{n=1}^{2^r}1/n\ge1+r/2$ for every $r\ge1$. The defining supremum is therefore $+\infty$.

No simple minorant has infinite integral. If it has any positive values, their finite set has a smallest positive value $c>0$. Wherever the minorant is positive, it is at least $c$; because it is bounded above by $1/n$, such points must satisfy $n\le1/c$. There are only finitely many such integers, and the minorant has finite values, so its counting-measure integral is finite. If it has no positive values, its integral is zero. Thus the supremum $+\infty$ is not attained.

Check: For $N=4$, the minorant has values $1,1/2,1/3,1/4$ at $1,2,3,4$, respectively, and zero afterward. Its integral is $1+1/2+1/3+1/4=25/12$, a finite value. The block $5\le n\le8$ contributes at least $4\cdot(1/8)=1/2$, as the general estimate states.

A Pitfall About Strict Improvement

Adding a positive amount on a set of positive finite measure always changes the value of a simple-function integral by the corresponding finite increment. But that does not always mean the new integral is strictly larger: extended-real arithmetic gives $\infty+c=\infty$ for finite $c$. This distinction is why the proposition required $\int\varphi\,d\mu<\infty$.

Worked Example: An Infinite Starting Integral

On $\mathbb{R}$ with Lebesgue measure $m$, let $f=2$, $\varphi=1$, and $E=[0,1]$. Then $\varphi\le f$, the gap $f-\varphi$ is $1$ on $E$, and $m(E)=1$. With $\varepsilon=1$, the raised minorant is $\psi=\varphi+\chi_E$, which equals $2$ on $[0,1]$ and $1$ elsewhere. Directly, \[ \int\varphi\,dm=m(\mathbb{R})=\infty,\qquad \int\psi\,dm=2m([0,1])+m(\mathbb{R}\setminus[0,1])=\infty. \] The refinement identity reads $\int\psi\,dm=\int\varphi\,dm+\varepsilon m(E)=\infty+1=\infty$. The increment is positive and finite, but the integral has not strictly increased. This example violates the finite-integral hypothesis of the proposition, so it does not contradict it.

Nonattainment is not a failure of approximation: in the interval example, explicit simple minorants approach the exact supremum as closely as desired. It is a statement about each individual simple function, not about the quality of the whole approximating family. Conversely, an infinite supremum need not be attained either, as counting measure demonstrates. Whenever strict improvement is used to rule out attainment, check both that the gap occurs on a positive-measure set and that the proposed minorant has finite integral.

Check Your Understanding

In each question, distinguish the value of the supremum from whether any particular simple minorant realizes it.

For $f(x)=x$ on $[0,1]$, why does the inclusion $\{\varphi>t\}\subseteq\{x>t\}$ hold for every simple minorant and every $t\ge0$?
Compute $\int\varphi_3\,dm$ for the dyadic minorant of $x$ on $[0,1]$, and compare it with $1/2$.
For $f=\chi_{[0,1/2]}$, identify a simple minorant attaining the supremum and justify the upper bound for all simple minorants.
Why does every simple minorant of $f(n)=1/n$ under counting measure have finite integral, even though their integrals have infinite supremum?
In the strict-improvement proposition, what goes wrong if $\int\varphi\,d\mu=\infty$, and which hypothesis ensures the claimed strict inequality?