The Essential Supremum

Upper Bounds That Ignore Null Sets

A measurable function can have a very large value on a set of measure zero without that value affecting its essential supremum. Conversely, values that occur on sets of positive measure cannot be discarded, however small those sets are. The essential supremum records this distinction: it is the least upper bound that holds almost everywhere, not necessarily the largest value the function takes at any point.

Throughout, let \((X,\mathcal{M},\mu)\) be any measure space; no finiteness, sigma-finiteness, or completeness assumption is imposed. Unless stated otherwise, \(f:X\to\mathbb{R}\) is real-valued and measurable, and statements involving \(f\le a\) almost everywhere mean that there is a measurable null set outside which the inequality holds. For a measurable set \(E\), “null” means \(\mu(E)=0\).

Definition: A real number \(a\) is an almost-everywhere upper bound for \(f\) if \(f\le a\) \(\mu\)-a.e.; equivalently, \(\mu(\{x:f(x)>a\})=0\). Let \(U_f=\{a\in\mathbb{R}:\mu(\{f>a\})=0\}\). The essential supremum is \(\operatorname*{ess\,sup}_X f=\inf U_f\), with the conventions \(\inf\varnothing=+\infty\) and \(\inf\mathbb{R}=-\infty\). It may therefore take either extended-real value \(+\infty\) or \(-\infty\).

The strict inequality in the level set \(\{f>a\}\) is important: \(a\) is an admissible upper bound precisely when the values strictly above \(a\) occur only on a null set. The set \(U_f\) is upward closed. Indeed, if \(a\in U_f\) and \(b\ge a\), then \(\{f>b\}\subseteq\{f>a\}\), so \(b\in U_f\). The definition uses the infimum of all such bounds; it does not assume in advance that this infimum itself is an almost-everywhere upper bound.

The Threshold Characterization

Theorem: Let \(f:X\to\mathbb{R}\) be measurable on \((X,\mathcal{M},\mu)\), and write \(S=\operatorname*{ess\,sup}_X f\). If \(S\in\mathbb{R}\), then \(f\le S\) \(\mu\)-a.e., and for every real \(a<S\), \(\mu(\{f>a\})>0\). If \(S=+\infty\), then \(\mu(\{f>a\})>0\) for every real \(a\). If \(S=-\infty\), then \(\mu(X)=0\); conversely, if \(\mu(X)=0\), then \(S=-\infty\).

This gives a useful way to read the value of \(S\): a finite essential supremum is itself an almost-everywhere upper bound, while every strictly smaller threshold is exceeded on a set of positive measure. If the essential supremum is \(+\infty\), no finite threshold is an almost-everywhere upper bound.

Proof. First suppose \(S\) is finite. Then \(U_f\) is nonempty, since an empty set has infimum \(+\infty\). For each positive integer \(n\), the definition of infimum gives \(u_n\in U_f\) with \(u_n<S+1/n\). For every \(n\), the set \(N_n=\{f>u_n\}\) is measurable and null. The countable union \(N=\bigcup_{n=1}^{\infty}N_n\) is null. If \(x\notin N\), then \(f(x)\le u_n<S+1/n\) for every \(n\). If \(f(x)>S\), one could choose \(n\) so large that \(1/n<f(x)-S\), contradicting \(f(x)<S+1/n\). Thus \(f\le S\) outside \(N\).

Now let \(a<S\). If \(\mu(\{f>a\})=0\), then \(a\in U_f\), so \(\inf U_f\le a\), contradicting \(a<S\). Hence \(\mu(\{f>a\})>0\). If \(S=+\infty\), \(U_f\) must be empty: every member of a nonempty subset of \(\mathbb{R}\) would give that subset a finite upper bound on its infimum. Thus no real \(a\) belongs to \(U_f\), and each \(\{f>a\}\) has positive measure.

Finally, if \(S=-\infty\), then \(U_f\) is unbounded below. For each positive integer \(n\), choose \(u_n\in U_f\) with \(u_n<-n\). The inclusion \(\{f>-n\}\subseteq\{f>u_n\}\) shows that \(\{f>-n\}\) is null. Since \(f\) is real-valued, \(X=\bigcup_{n=1}^{\infty}\{f>-n\}\): for each \(x\), some positive integer \(n\) satisfies \(-n<f(x)\). Countable subadditivity gives \(\mu(X)=0\). Conversely, if \(\mu(X)=0\), every set \(\{f>a\}\) is null, so \(U_f=\mathbb{R}\) and \(\inf U_f=-\infty\). \(\square\)

Worked Example: A Step Function on an Interval

Let \(X=[0,1]\) with the Lebesgue sigma-algebra and Lebesgue measure \(m\). Define \(f:[0,1]\to\mathbb{R}\) by \(f=5\) on \([0,1/4)\), \(f=3\) on \([1/4,3/4)\), and \(f=-1\) on \([3/4,1]\). Compute its essential supremum.

For any real \(a<5\), every point of \([0,1/4)\) satisfies \(f(x)=5>a\). Therefore \(m(\{f>a\})\ge m([0,1/4))=1/4>0\), and \(a\) is not an almost-everywhere upper bound. For every \(a\ge5\), there are no points with \(f(x)>a\), so \(m(\{f>a\})=0\). Thus \(U_f=[5,\infty)\), and

\operatorname*{ess\,sup}_{[0,1]}f=\inf[5,\infty)=5.

Check: \(f\le5\) everywhere, and every threshold strictly below 5 is exceeded on a set of measure at least \(1/4\). This verifies both parts of the finite threshold characterization.

Worked Example: A Large Value on a Null Set

On \([0,1]\) with Lebesgue measure, define \(f(x)=x\) for \(x\ne0\) and \(f(0)=100\). The function is measurable because it differs from the continuous function \(x\mapsto x\) only on the measurable singleton \(\{0\}\). Find its essential supremum.

If \(a\ge1\), then \(\{f>a\}\) is either empty or contained in \(\{0\}\): on \((0,1]\), \(f(x)=x\le1\). In either case, \(m(\{f>a\})=0\). If \(0\le a<1\), the interval \((a,1]\) lies in \(\{f>a\}\), so \(m(\{f>a\})=1-a>0\). If \(a<0\), all of \((0,1]\) lies in \(\{f>a\}\), so the level set has measure at least 1. Therefore the admissible upper bounds are exactly \([1,\infty)\), and \(\operatorname*{ess\,sup}_{[0,1]}f=1\).

Check: \(f\le1\) except at \(x=0\), a null set, and for every \(a<1\) the level set above \(a\) has positive measure. The pointwise supremum is 100, but the essential supremum is 1.

Worked Example: A Finite Essential Supremum That Is Not Attained

Give \(\mathbb{N}=\{1,2,\ldots\}\) counting measure, for which the measure of a set is its number of elements (possibly \(+\infty\)). Let \(f(n)=1-1/n\). Every nonempty set has positive counting measure, so an almost-everywhere upper bound here must be an upper bound at every \(n\).

For every \(n\), \(f(n)=1-1/n<1\), so 1 is an upper bound. If \(0\le a<1\), choose an integer \(n>1/(1-a)\); then \(1/n<1-a\), hence \(f(n)=1-1/n>a\). If \(a<0\), \(f(1)=0>a\). Consequently every \(a<1\) fails to be an upper bound and every \(a\ge1\) is an upper bound. Thus \(U_f=[1,\infty)\) and \(\operatorname*{ess\,sup}_{\mathbb{N}}f=1\).

Check: \(f(n)<1\) for each individual \(n\), but the values approach 1: \(f(n)=1-1/n\to1\). The essential supremum need not be attained at any point.

Worked Example: An Infinite Essential Supremum

On \(X=(0,1)\) with Lebesgue measure, let \(f(x)=1/x\). For \(a\ge1\), the strict level set is \(\{f>a\}=(0,1/a)\), whose measure is \(1/a>0\). For \(a<1\), every \(x\in(0,1)\) satisfies \(1/x>1>a\), so \(\{f>a\}=(0,1)\) and its measure is 1. No finite \(a\) is an almost-everywhere upper bound. Therefore \(U_f=\varnothing\) and \(\operatorname*{ess\,sup}_{(0,1)}f=+\infty\).

Check: the formula for \(a\ge1\) uses the strict condition \(1/x>a\), equivalent to \(x<1/a\); the endpoint \(x=1/a\) is not included and does not affect the measure.

Null-Set Invariance and a Common Pitfall

Proposition: Suppose \(f,g:X\to\mathbb{R}\) are measurable on \((X,\mathcal{M},\mu)\), and \(f=g\) \(\mu\)-a.e. Then \(\operatorname*{ess\,sup}_X f=\operatorname*{ess\,sup}_X g\), including when the common value is \(+\infty\) or \(-\infty\).

Proof. Since \(f=g\) \(\mu\)-a.e., there is a measurable null set \(N\) such that \(f(x)=g(x)\) for all \(x\notin N\). For every real \(a\), the symmetric difference \(\{f>a\}\triangle\{g>a\}\) is contained in \(N\). A set differs from a null set modification by at most a null set, so \(\mu(\{f>a\})=0\) if and only if \(\mu(\{g>a\})=0\). Thus \(U_f=U_g\), and their infima, with the stated conventions, are equal. \(\square\)

A common mistake is to calculate the essential supremum by taking the pointwise supremum and then stopping. For the function in the null-set example, that calculation gives \(\sup_{x\in[0,1]}f(x)=100\), and the incorrect conclusion would be \(\operatorname*{ess\,sup}f=100\). The correction is to examine \(\{f>a\}\): for every \(a\ge1\), this set is contained in the measure-zero singleton \(\{0\}\), whereas every \(a<1\) is exceeded on a set of positive measure. This level-set test gives the correct value, 1.

The proposition explains why the essential supremum depends only on the almost-everywhere equivalence class of a measurable function. This is the relevant viewpoint for \(L^p\) spaces: functions equal almost everywhere represent the same element. At this stage, the essential supremum provides the upper-bound notion used to describe the endpoint \(p=\infty\); calculating the associated \(L^\infty\) norm requires keeping track of this same null-set distinction.

Check Your Understanding

For each question, use the strict level sets \(\{f>a\}\) and state whether the relevant measure is Lebesgue measure or counting measure.

On \([0,1]\) with Lebesgue measure, let \(f=2\) on \([0,1/3)\) and \(f=-4\) on \([1/3,1]\). Determine \(U_f\) and the essential supremum.
On \([0,1]\) with Lebesgue measure, let \(f(x)=x\) except that \(f(1)=20\). Does this modification change the essential supremum? Explain using level sets.
On an arbitrary measure space, suppose a real-valued measurable \(f\) satisfies \(\mu(\{f>a\})=0\) for every real \(a\). What does this imply about \(\mu(X)\) and the essential supremum?
On \(\mathbb{N}\) with counting measure, compute the essential supremum of \(g(n)=2+1/n\). Is it attained?
For a real-valued measurable \(f\) with finite essential supremum \(S\), why must \(\mu(\{f>a\})>0\) for every \(a<S\)?

What You'll Learn