Two Ways to Measure an Upper Bound
The pointwise supremum records every value of a function, including values at individual points. The essential supremum ignores values confined to sets of measure zero. This distinction matters in \(L^\infty\): its norm depends on the function up to equality almost everywhere, so changing a representative at even one point must not change the norm.
Let \((X,\mathcal{M},\mu)\) be any measure space; no finiteness, completeness, or sigma-finiteness assumption is needed. In the definitions and results below, \(u:X\to[0,\infty]\) is measurable. Its pointwise supremum is \(\sup_{x\in X}u(x)\), with value \(0\) if \(X\) is empty. An almost-everywhere upper bound \(a\ge0\) means \(\mu(\{x:u(x)>a\})=0\). The strict inequality in this level set is important: points where \(u(x)=a\) do not violate the bound.
The restriction \(a\ge0\) is part of this definition. It handles degenerate spaces correctly: if \(\mu(X)=0\), every nonnegative \(a\) is an almost-everywhere upper bound, and the essential supremum of any nonnegative \(u\) is \(0\). In particular, claims about positive-measure level sets below will only concern thresholds \(a\) with \(0\le a<s\), where \(s\) is the essential supremum. They make no claim about negative thresholds or about thresholds below \(s\) when \(s=0\).
How to Recognize the Essential Supremum
Proof. Write \(U=\{a\ge0:\mu(\{u>a\})=0\}\). If \(s<\infty\), then \(U\) is nonempty and \(s=\inf U\). For each positive integer \(n\), the definition of the infimum gives an \(a_n\in U\) with \(a_n<s+1/n\). Since \(a_n\in U\), we have \(u\le a_n<s+1/n\) outside the null set \(\{u>a_n\}\). Thus \(\{u>s+1/n\}\subseteq\{u>a_n\}\) is null. Every point where \(u>s\) belongs to \(\{u>s+1/n\}\) for some \(n\): choose \(n\) so large that \(1/n<u-s\) at that point. Hence \(\{u>s\}\subseteq\bigcup_{n=1}^\infty\{u>s+1/n\}\), a countable union of null sets. This proves \(u\le s\) a.e.
If \(0\le a<s\) and \(\mu(\{u>a\})=0\), then \(a\in U\), which would imply \(s=\inf U\le a\), a contradiction. So \(\mu(\{u>a\})>0\). Finally, if \(s=\infty\) but \(\mu(\{u>a\})=0\) for some finite \(a\ge0\), then \(a\in U\) and \(s\le a<\infty\), again a contradiction. \(\square\)
This characterization says that an essential supremum is a sharp almost-everywhere upper bound. When it is finite, the function stays at or below it almost everywhere, while every strictly smaller nonnegative threshold is exceeded on a set of positive measure. It does not say that \(\{u>s\}\) has positive measure; in fact, that set is null when \(s<\infty\).
Worked Example: A Single Exceptional Value
On \(X=[0,1]\) with the Lebesgue sigma-algebra \(\mathcal{L}\) and Lebesgue measure \(m\), let \(u(x)=5\) at \(x=1/2\) and \(u(x)=0\) otherwise. Compute its pointwise supremum and essential supremum.
The value \(5\) occurs at \(1/2\), so \(\sup_{x\in[0,1]}u(x)=5\). For any \(a\) with \(0\le a<5\), the strict superlevel set is \(\{u>a\}=\{1/2\}\), which has Lebesgue measure \(0\). Thus every such \(a\) is an almost-everywhere upper bound, and the infimum of the nonnegative upper bounds is \(0\). Therefore \(\operatorname*{ess\,sup}u=0\). At \(a=5\), the strict superlevel set is empty, so the endpoint also satisfies the bound. Check: \(u=0\) outside a singleton of measure \(0\), so \(u=0\) a.e., consistent with essential supremum \(0\).
Worked Example: A Supremum Approached on Positive-Measure Sets
On \([0,1]\) with Lebesgue measure, let \(u(x)=x\). Find both suprema and verify the level-set criterion.
The values satisfy \(x\le1\) throughout \([0,1]\), and \(u(1)=1\), so the pointwise supremum is \(1\), attained at \(x=1\). For any \(a\in[0,1)\), \(\{u>a\}=(a,1]\), whose measure is \(1-a>0\). For example, at \(a=3/4\), the set is \((3/4,1]\) and has measure \(1-3/4=1/4\). At \(a=1\), the strict superlevel set \(\{u>1\}\) is empty. Hence the essential supremum is also \(1\). Check: \(u\le1\) everywhere, and every nonnegative \(a<1\) fails to be an almost-everywhere upper bound because its strict superlevel set has measure \(1-a>0\).
Worked Example: An Unbounded Function That Is Zero Almost Everywhere
On \(\mathbb{R}\) with the Borel sigma-algebra and Lebesgue measure \(m\), define \(u(n)=n\) for positive integers \(n\), and \(u(x)=0\) for all other \(x\). Compute its suprema.
The function is measurable: its nonzero level sets are subsets of the countable Borel set \(\mathbb{N}\). Since \(u(n)=n\) for every positive integer \(n\), its pointwise supremum is \(+\infty\). For any finite \(a\ge0\), \(\{u>a\}=\{n\in\mathbb{N}:n>a\}\). This is countable, so its Lebesgue measure is \(0\). Thus every finite nonnegative \(a\) is an almost-everywhere upper bound and \(\operatorname*{ess\,sup}u=0\). Check: \(u=0\) outside \(\mathbb{N}\), and \(m(\mathbb{N})=0\), so \(u=0\) a.e. despite being unbounded pointwise.
Null-Set Changes and the \(L^\infty\) Norm
For measurable functions \(f,g:X\to\mathbb{C}\), equality \(\mu\)-a.e. means that \(\mu(\{x:f(x)\ne g(x)\})=0\). If \(f=g\) a.e., then for every \(a\ge0\), the symmetric difference of the level sets \(\{|f|>a\}\) and \(\{|g|>a\}\) is contained in \(\{f\ne g\}\). Consequently, one of these level sets is null exactly when the other is null, and their essential suprema agree.
The norm is defined from \(|f|\), a nonnegative measurable function. It is invariant under a.e. changes, so it is well-defined on equivalence classes rather than on particular pointwise representatives. The essential-supremum proposition also gives the threshold test: for \(c\ge0\), \(\|f\|_\infty\le c\) exactly when \(|f|\le c\) a.e. In one direction, \(\mu(\{|f|>c\})=0\) makes \(c\) an admissible bound and therefore \(\|f\|_\infty\le c\). In the other direction, if \(\|f\|_\infty\le c\), then for every \(b>c\), the proposition gives \(|f|\le b\) a.e.; taking \(b=c+1/n\) for each \(n\) and removing the countable union of exceptional null sets gives \(|f|\le c\) a.e.
In particular, \(\|f\|_\infty=0\) if and only if \(f=0\) a.e. If the norm is zero, the threshold test with \(c=0\) gives \(|f|=0\) a.e. Conversely, if \(f=0\) a.e., then \(c=0\) is an admissible bound, so the infimum defining the norm is \(0\). This explains precisely why a pointwise supremum cannot replace the essential supremum in \(L^\infty\): it can change when a representative is altered on a null set.
Worked Example: Changing One Representative
On \([0,1]\) with Lebesgue measure, let \(f(x)=2\) except at \(x=1/3\), where \(f(1/3)=100\). Compare the pointwise supremum with the \(L^\infty\) norm.
The pointwise supremum is \(100\). If \(0\le c<2\), then \(\{f>c\}=[0,1]\), which has measure \(1\). If \(c=2\), then \(\{f>2\}=\{1/3\}\), which has measure \(0\). Therefore the least nonnegative almost-everywhere upper bound is \(2\), and \(\|f\|_\infty=2\). Check: changing \(f(1/3)\) from \(100\) to \(2\) gives the constant function \(g=2\), with pointwise supremum \(2\) and the same \(L^\infty\) norm \(2\); indeed \(f=g\) a.e.
A Common Mistake and Its Correction
A common mistake is to identify the essential supremum with the pointwise supremum, or to claim that a function is not essentially bounded just because it takes arbitrarily large values. For the function on \(\mathbb{R}\) in the third example, one might reason, “\(u(n)=n\), so \(\operatorname*{ess\,sup}u=+\infty\).” That calculation confuses the pointwise supremum with the essential one. For every fixed finite threshold \(a\ge0\), the strict superlevel set \(\{u>a\}\) is countable and has measure zero. The corrected calculation is therefore \(\operatorname*{ess\,sup}u=0\), while \(\sup_x u(x)=+\infty\).
The reverse warning is also useful: essential supremum does not discard large values merely because they occur on small sets. On \((0,1)\) with Lebesgue measure, \(v(x)=1/x\) has \(\{v>K\}=(0,1/K)\) for each \(K\ge1\), and this set has measure \(1/K>0\). Thus no finite \(K\) is an almost-everywhere upper bound, so \(\operatorname*{ess\,sup}v=+\infty\). The relevant distinction is null set versus positive measure, not large set versus small set.
Check Your Understanding
For each question, distinguish a pointwise statement from an almost-everywhere one and use strict superlevel sets where appropriate.
- On \([0,1]\) with Lebesgue measure, let \(u=7\) at one point and \(u=1\) elsewhere. Find the pointwise supremum and the essential supremum.
- Suppose \(u:X\to[0,\infty]\) is measurable and \(\operatorname*{ess\,sup}u=s<\infty\). What does the definition imply about \(\mu(\{u>a\})\) for \(0\le a<s\), and about \(\mu(\{u>s\})\)?
- On \(\mathbb{R}\) with Lebesgue measure, define \(u(n)=1\) for \(n\in\mathbb{N}\) and \(u=0\) elsewhere. Compute \(\sup_xu(x)\) and \(\operatorname*{ess\,sup}u\).
- Why does changing a complex-valued measurable function at one point leave its \(L^\infty\) norm unchanged when that point has measure zero?
- On \((0,1)\) with Lebesgue measure, show from the strict superlevel sets that \(v(x)=1/x\) has infinite essential supremum.