From Essential Bounds to the \(L^\infty\) Norm
The ordinary supremum records every value a function takes, including values at isolated points or on other null sets. The \(L^\infty\) norm ignores such changes: it measures the smallest bound that holds almost everywhere. Computing it therefore requires checking the measure of strict level sets, not just finding a largest pointwise value.
Let \((X,\mathcal{M},\mu)\) be any measure space; no finiteness, sigma-finiteness, or completeness assumption is needed here. Functions in this discussion are measurable and real-valued or complex-valued. Two functions \(f\) and \(g\) are identified if \(f=g\) \(\mu\)-a.e.; that is, if \(\mu(\{x:f(x)\ne g(x)\})=0\). The space \(L^\infty(\mu)\) consists of these equivalence classes that have a finite almost-everywhere bound.
The strict inequality in the level set is important. A bound \(K\) is an essential bound when \(|f|\le K\) outside a null set, equivalently when \(\mu(\{|f|>K\})=0\). In particular, the values on the level set \(\{|f|=K\}\) do not violate the bound. The definition also covers a measure space of measure zero: then every function is zero almost everywhere, and its norm is \(0\).
- if \(K\ge M\), then \(\mu(\{|f|>K\})=0\);
- if \(K<M\), then \(\mu(\{|f|>K\})>0\).
Proof. Write \(S=\{K\ge0:\mu(\{|f|>K\})=0\}\). If \(S\) is empty, its infimum is \(+\infty\), and the claimed positive-measure statement for every finite \(K\) follows directly from the definition of \(S\). Now suppose \(S\) is nonempty and \(M=\inf S<\infty\). For each positive integer \(n\), the definition of infimum gives a \(K_n\in S\) with \(K_n<M+1/n\). Outside the null set \(N=\bigcup_{n=1}^\infty\{|f|>K_n\}\), we have \(|f|\le K_n<M+1/n\) for every \(n\). If \(|f(x)|>M\) at some \(x\notin N\), choosing \(n\) so large that \(1/n<|f(x)|-M\) would contradict \(|f(x)|<M+1/n\). Thus \(\{|f|>M\}\subseteq N\), so it is null. If \(K\ge M\), then \(\{|f|>K\}\subseteq\{|f|>M\}\), giving the first case. If \(0\le K<M\) and \(\{|f|>K\}\) were null, then \(K\in S\), contradicting \(M=\inf S>K\). This proves both cases. \(\square\)
This test gives a practical rule: to show that a proposed value \(M\) is the norm, show that the set where \(|f|>M\) is null and that \(\{|f|>K\}\) has positive measure for every \(0\le K<M\). When \(M=0\), only the first check is needed.
Continuous Functions on Compact Intervals
Proof. By the Weierstrass extreme value theorem, the continuous function \(|f|\) has a maximum \(M\) at some \(x_0\in[a,b]\). Since \(|f(x)|\le M\) everywhere, the set \(\{|f|>M\}\) is empty, so \(M\) is an essential bound. If \(M=0\), this already gives \(\|f\|_\infty=0\). Suppose \(M>0\), and fix \(K\) with \(0\le K<M\). By continuity at \(x_0\), there is a \(\delta>0\) such that \(|f(x)|>K\) whenever \(x\in[a,b]\) and \(|x-x_0|<\delta\). This relative neighborhood has positive Lebesgue measure: since \(a<b\), its intersection with \([a,b]\) contains an interval of positive length, including when \(x_0\) is an endpoint. Therefore \(m(\{|f|>K\})>0\). The level-set test now gives \(\|f\|_\infty=M\). \(\square\)
Worked Example: A Continuous Function Whose Maximum Is at an Endpoint
On \(X=[0,1]\) with its Lebesgue sigma-algebra and Lebesgue measure \(m\), let \(f(x)=2x-x^2\). Compute \(\|f\|_\infty\), taking account of the endpoint where the maximum occurs.
Rewrite \(f(x)=1-(1-x)^2\). Because \(0\le1-x\le1\), we have \(0\le f(x)\le1\), and \(f(1)=1\). Thus the pointwise maximum is \(1\). To verify it is also the essential supremum, fix \(K\) with \(0\le K<1\). The inequality \(f(x)>K\) is equivalent to $$ 1-(1-x)^2>K \quad\Longleftrightarrow\quad (1-x)^2<1-K \quad\Longleftrightarrow\quad x>1-\sqrt{1-K}. $$ Consequently, \(\{x\in[0,1]:f(x)>K\}=(1-\sqrt{1-K},1]\), whose measure is \(\sqrt{1-K}>0\). At \(K=1\), the strict level set \(\{f>1\}\) is empty. The level-set test gives \(\|f\|_\infty=1\). Check: at \(K=0\), the formula gives a level-set measure of \(\sqrt{1-0}=1\), consistent with \(f(x)>0\) exactly on \((0,1]\); at \(K=1\), the measure is \(0\).
Finite Step Functions
Proof. Let the displayed maximum be \(M\). If \(\mu(E_j)=0\), that piece contributes only a null set; every other piece has \(|c_j|\le M\). Since there are finitely many pieces, \(\{|f|>M\}\) is contained in the finite union of the null pieces, and is null. If \(M>0\), some \(E_j\) has positive measure and \(|c_j|=M\). For every \(0\le K<M\), that entire \(E_j\) lies in \(\{|f|>K\}\), so this level set has positive measure. The level-set test gives \(\|f\|_\infty=M\). If there are no positive-measure pieces, their finite union \(X\) is null; then \(f=0\) a.e. and \(\|f\|_\infty=0=M\). \(\square\)
Worked Example: A Step Function with Three Pieces
On \(X=[0,4)\) with Lebesgue measure \(m\), define \(f=-2\) on \(E_1=[0,1)\), \(f=3\) on \(E_2=[1,5/2)\), and \(f=1/2\) on \(E_3=[5/2,4)\). Compute its norm using the pieces and verify it with strict level sets.
The piece measures are \(m(E_1)=1-0=1\), \(m(E_2)=5/2-1=3/2\), and \(m(E_3)=4-5/2=3/2\). All three are positive. The absolute values of the corresponding function values are \(|-2|=2\), \(|3|=3\), and \(|1/2|=1/2\), so the largest is \(3\). For any \(0\le K<3\), all of \(E_2\) satisfies \(|f|=3>K\); hence \(m(\{|f|>K\})\ge m(E_2)=3/2>0\). At \(K=3\), no piece has \(|f|>3\), so \(m(\{|f|>3\})=0\). Therefore \(\|f\|_\infty=3\). Check: the formula for the positive-measure pieces gives \(\max\{2,3,1/2\}=3\), agreeing with the level-set calculation.
Changing a Function on a Null Set
If two measurable functions agree \(\mu\)-a.e., then their strict level sets \(\{|f|>K\}\) and \(\{|g|>K\}\) can differ only on a null set, for each \(K\ge0\). Thus one of these sets is null exactly when the other is null, and their essential suprema are equal. This is why the \(L^\infty\) norm belongs to the almost-everywhere equivalence class, not to a particular choice of values on a null set.
Worked Example: Changing Values at Every Rational Point
On \(X=[0,1]\) with Lebesgue measure \(m\), define \(g(x)=x\) when \(x\) is irrational and \(g(x)=7\) when \(x\) is rational. Compute the pointwise supremum and the \(L^\infty\) norm separately.
The rationals \(Q\cap[0,1]\) form a countable set, so they are Borel and have Lebesgue measure zero. The function \(g\) is measurable: it equals the continuous function \(x\) off this Borel null set and is constant on the set. Its pointwise supremum is \(7\), since rational points occur in \([0,1]\) and \(g=7\) there. For \(0\le K<1\), the irrational points \(x\in(K,1]\) satisfy \(g(x)=x>K\). Their measure is \(1-K\), because deleting the rationals removes a null set. Hence \(m(\{g>K\})\ge1-K>0\). At \(K=1\), \(\{g>1\}=Q\cap[0,1]\), which has measure zero. Therefore \(\|g\|_\infty=1\). Check: \(g=x\) except on a null set, and \(x\) has norm \(1\) on \([0,1]\); the level-set calculation gives the same result while the pointwise supremum is \(7\).
When No Finite Essential Bound Exists
Worked Example: A Finite-Valued Function with Infinite \(L^\infty\) Norm
On \(X=(0,1]\) with Lebesgue measure \(m\), let \(h(x)=x^{-1/2}\). Show that \(h\) is finite at every point but does not belong to \(L^\infty(m)\).
For each finite \(K>0\), the strict inequality \(x^{-1/2}>K\) is equivalent to \(x<1/K^2\). Thus $$ m(\{x\in(0,1]:h(x)>K\})= \begin{cases} 1,&0<K\le1,\\ 1/K^2,&K>1. \end{cases} $$ In both cases this measure is positive. At \(K=0\), the level set is all of \((0,1]\), also of measure \(1\). No finite \(K\) is an essential bound, so \(\operatorname*{ess\,sup}h=+\infty\), and \(h\notin L^\infty(m)\). Check: for \(K=2\), the set is \((0,1/4)\), with measure \(1/4=1/2^2\), as the formula predicts.
A Common Mistake
A tempting but incorrect calculation for the rational modification above is: “\(g\) takes the value \(7\), so \(\|g\|_\infty=7\).” This computes the pointwise supremum, not the \(L^\infty\) norm. The correction is to test the sets \(\{|g|>K\}\): for \(K=1\), the set consists only of rationals and has measure zero, while for every \(K<1\), it contains irrational points in \((K,1]\) and has positive measure. Therefore the essential supremum is \(1\). The same distinction matters for step functions: a large value on a piece of measure zero does not affect the norm.
For a proposed finite answer \(M\), the decisive checks are \(\mu(\{|f|>M\})=0\) and, when \(M>0\), \(\mu(\{|f|>K\})>0\) for every \(0\le K<M\). The first confirms that \(M\) is an essential bound; the second rules out every smaller one. These tests work on arbitrary measure spaces and make explicit which values are ignored and why.
Check Your Understanding
For each question, distinguish pointwise values from almost-everywhere bounds and use strict level sets where appropriate.
- On \([0,1]\) with Lebesgue measure, compute the \(L^\infty\) norm of \(f(x)=1-x\). For \(0\le K<1\), find the measure of \(\{f>K\}\).
- A measurable step function equals \(4\) on a null set and \(-2\) on a set of positive measure, and is \(0\) elsewhere. What is its \(L^\infty\) norm, and why does the value \(4\) not determine it?
- Suppose \(f=g\) \(\mu\)-a.e. on a measure space. Explain why their strict level sets have the same measure-zero status for each \(K\ge0\).
- On \((0,1]\) with Lebesgue measure, verify directly that \(\{x:x^{-1}>K\}\) has positive measure for every finite \(K\ge0\). What does this imply about the \(L^\infty\) norm?
- State the two level-set checks that establish a proposed positive value \(M\) as an essential supremum.