First Check What “Simple Functions” Means
Claims about density and inclusion often fail because a valid theorem is applied with its direction reversed, or because an unstated restriction on the approximating functions is overlooked. The \(L^\infty\) case illustrates the second problem: with the standard definition of a measurable simple function, the claim that simple functions are dense in \(L^\infty\) is true. A related claim becomes false if “simple functions” is restricted to functions supported on sets of finite measure.
Throughout, \((X,\mathcal{M},\mu)\) is a measure space. Unless specified otherwise, functions are real-valued and measurable, and \(L^p(\mu)\) denotes equivalence classes modulo equality \(\mu\)-a.e. For \(1\le p<\infty\), \(L^p(\mu)\) consists of functions with \(\int_X |f|^p\,d\mu<\infty\), with norm \(\|f\|_p=(\int_X|f|^p\,d\mu)^{1/p}\). The space \(L^\infty(\mu)\) consists of functions whose essential supremum \(\|f\|_\infty\) is finite. A simple function here means a measurable function taking only finitely many values; its level sets need not have finite measure.
Simple Functions Are Dense in \(L^\infty\)—With the Usual Definition
Let \(f\in L^\infty(\mu)\). Choose \(C>\|f\|_\infty\). Then \(|f|\le C\) outside a measurable null set \(N\). Replacing \(f\) by zero on \(N\) does not change its \(L^\infty\) equivalence class, and gives a measurable representative bounded everywhere by \(C\). Divide the interval \([-C,C]\) into finitely many measurable intervals of length at most \(\delta\), and on the inverse image of each interval replace \(f\) by one endpoint. The resulting function is a finite-valued measurable simple function \(s\), and \(|f-s|\le\delta\) everywhere for the chosen representative. Thus \(\|f-s\|_\infty\le\delta\). Since \(\delta\) can be arbitrarily small, unrestricted measurable simple functions are dense in \(L^\infty(\mu)\). This argument needs no finiteness or sigma-finiteness assumption on \(\mu\).
Worked Example: Quantizing a Bounded Function
Take \(X=[0,1]\), with Lebesgue measure \(m\), and \(f(x)=x\). For each positive integer \(n\), define \(s_n(x)=k/n\) when \(k/n\le x<(k+1)/n\), for \(k=0,\ldots,n-1\), and set \(s_n(1)=1\). The function \(s_n\) is measurable and takes the \(n+1\) values \(0,1/n,\ldots,1\).
On each interval \([k/n,(k+1)/n)\), the error is \(x-k/n\), so \(0\le f(x)-s_n(x)<1/n\). At \(x=1\), the error is zero. Hence \(\|f-s_n\|_\infty\le1/n\). To verify that this norm is exactly \(1/n\), fix any \(c<1/n\) with \(c\ge0\). On the interval \((1-1/n+\eta,1)\), where \(\eta\) is chosen with \(0<\eta<1/n-c\), the error exceeds \(c\) on a set of positive measure. Therefore the essential supremum is at least \(c\) for every such \(c\), and so \(\|f-s_n\|_\infty=1/n\).
Check: Substituting \(n=4\) gives \(\|f-s_4\|_\infty=1/4\). The values of \(s_4\) on the four half-open intervals are \(0,1/4,1/2,3/4\), and the error approaches \(1/4\) on each interval from its right side. At the single point \(x=1\) the error is zero, which does not change the essential supremum.
A common restriction changes the conclusion. Suppose \(\mu(X)=\infty\) and approximants are required to vanish outside a measurable set of finite measure. Such a simple function cannot approximate the constant function \(1\) in \(L^\infty\): if it vanishes outside \(A\) with \(\mu(A)<\infty\), then \(\mu(X\setminus A)=\infty\), and \(|1-s|=1\) throughout \(X\setminus A\). Its essential-supremum distance from \(1\) is therefore at least \(1\). The distinction is not “simple functions versus no simple functions”; it is unrestricted simple functions versus a restricted approximating class.
Worked Example: Finite-Measure Support Does Not Approximate 1 on the Line
Let \(X=\mathbb{R}\), \(\mathcal{M}=\mathcal{L}\), and let \(\mu=m\), Lebesgue measure. The constant function \(f(x)=1\) belongs to \(L^\infty(m)\), with \(\|f\|_\infty=1\). Let \(s\) be any measurable simple function that vanishes outside some measurable \(A\) with \(m(A)<\infty\). Since \(m(\mathbb{R})=\infty\), \(m(\mathbb{R}\setminus A)=\infty\); otherwise \(\mathbb{R}=A\cup(\mathbb{R}\setminus A)\) would have finite measure. On \(\mathbb{R}\setminus A\), \(|f-s|=|1-0|=1\). Consequently \(\|f-s\|_\infty\ge1\), so no such \(s\) is within, for example, \(1/2\) of \(f\). The value \(1\) is also an upper bound on the distance for the zero function \(s=0\), so the infimum of these distances is exactly \(1\).
Check: For \(s=0\), \(|f-s|=1\) everywhere, giving \(\|f-s\|_\infty=1\), consistent with the lower bound for every finite-measure-support approximant.
Finite Measure Gives an Embedding in One Direction
On a finite-measure space, the direction of the \(L^p\) inclusion is from the larger exponent to the smaller one. An embedding here means inclusion with a norm bound.
Proof. First suppose \(q<\infty\), and take \(f\in L^q(\mu)\). Hölder's inequality, with conjugate exponents \(q/p\) and \(q/(q-p)\), applies because \(q/p>1\) and \((q/p)^{-1}+(q/(q-p))^{-1}=1\). Applying it to \(|f|^p\) and the constant function \(1\) gives $$ \int_X |f|^p\,d\mu \le \left(\int_X |f|^q\,d\mu\right)^{p/q} \left(\int_X 1\,d\mu\right)^{1-p/q} =\|f\|_q^p M^{1-p/q}. $$ The right side is finite. Taking the \(p\)-th root gives the stated bound. If \(q=\infty\), then \(|f|\le\|f\|_\infty\) \(\mu\)-a.e., so $$ \int_X|f|^p\,d\mu\le \|f\|_\infty^p\mu(X)=\|f\|_\infty^pM. $$ Taking the \(p\)-th root proves the second bound and the inclusion. If \(M=0\), every measurable set has measure zero, so every measurable function is zero a.e.; this proves the final case. \(\square\)
Hölder's inequality in the proof is the standard result from the earlier material on \(L^p\) spaces. The finite-measure hypothesis enters through the factor \(\int_X1\,d\mu=M\); it is not a harmless detail. Also, the theorem does not assert the reverse inclusion. The next example shows why reversing it can produce an immediate contradiction.
Worked Example: The Reverse Finite-Measure Inclusion Fails
Take \(X=(0,1)\) with Lebesgue measure, and define \(f(x)=x^{-3/4}\) for \(0<x<1\). The space has measure \(1\). For the \(L^1\) integral, direct integration gives $$ \int_0^1|f(x)|\,dm(x)=\int_0^1x^{-3/4}\,dx =\lim_{\varepsilon\downarrow0}\left[4x^{1/4}\right]_{\varepsilon}^{1} =\lim_{\varepsilon\downarrow0}4(1-\varepsilon^{1/4})=4. $$ Thus \(f\in L^1(m)\) and \(\|f\|_1=4\). For the square, integration over \((\varepsilon,1)\) gives $$ \int_\varepsilon^1|f(x)|^2\,dm(x) =\int_\varepsilon^1x^{-3/2}\,dx =2(\varepsilon^{-1/2}-1), $$ which tends to \(+\infty\) as \(\varepsilon\downarrow0\). Hence \(f\notin L^2(m)\). So even on a space of measure \(1\), \(L^1\subseteq L^2\) is false in general; the guaranteed inclusion goes the other way.
Check: The constant function \(g(x)=1\) satisfies \(\|g\|_1=1\) and \(\|g\|_2=1\). This agrees with the theorem's \(L^2\)-to-\(L^1\) bound, whose constant here is \(1^{1-1/2}=1\), and does not repair the failed reverse inclusion for \(f\).
Infinite Measure Removes Both General Finite-\(p\) Inclusions
When \(X\) has infinite measure, there is no general inclusion in either direction between \(L^p(\mu)\) and \(L^q(\mu)\) for distinct finite exponents \(1\le p<q<\infty\). The finite-measure estimate cannot be used: its factor involving \(\mu(X)\) is no longer finite. For Lebesgue measure on \(\mathbb{R}\), the following pair shows both failures already for \(p=1\) and \(q=2\).
Worked Example: Both Directions Fail on the Real Line
Let \(X=\mathbb{R}\), with Lebesgue measure \(m\). Define \(f(x)=x^{-3/4}\) for \(0<x<1\) and \(f(x)=0\) elsewhere. The value at \(x=0\) may be set to zero and does not affect the integrals. As computed above, \(\int_{\mathbb{R}}|f|\,dm=4\), while \(\int_{\mathbb{R}}|f|^2\,dm=\infty\). Thus \(f\in L^1(m)\) but \(f\notin L^2(m)\).
Now define \(h(x)=x^{-3/4}\) for \(x\ge1\) and \(h(x)=0\) for \(x<1\). Its \(L^1\) integral, evaluated up to \(R>1\), is $$ \int_1^R x^{-3/4}\,dx=4(R^{1/4}-1), $$ which tends to \(+\infty\) as \(R\to\infty\). On the other hand, $$ \int_{\mathbb{R}}|h|^2\,dm =\int_1^\infty x^{-3/2}\,dx =\lim_{R\to\infty}\left[-2x^{-1/2}\right]_1^R =\lim_{R\to\infty}2(1-R^{-1/2})=2. $$ Therefore \(h\in L^2(m)\) with \(\|h\|_2=\sqrt{2}\), but \(h\notin L^1(m)\).
Check: The two computations have opposite outcomes: \(f\) rules out \(L^1(m)\subseteq L^2(m)\), and \(h\) rules out \(L^2(m)\subseteq L^1(m)\). The space \(\mathbb{R}\) has infinite measure, so the finite-measure embedding theorem makes neither claim.
For any fixed \(1\le p<q<\infty\), the same distinction can be arranged with powers: near zero, choose \(a\) with \(1/q\le a<1/p\), so \(x^{-a}\) is \(p\)-integrable but not \(q\)-integrable on \((0,1)\). On \([1,\infty)\), choose \(b\) with \(1/q<b\le1/p\), so \(x^{-b}\) is \(q\)-integrable but not \(p\)-integrable. These choices are possible because \(p<q\). The explicit \(p=1,q=2\) calculations above are instances of these strict power thresholds.
A Pitfall: Reversing the Norm Inequality
A tempting but incorrect calculation on a probability space, or any space of measure \(1\), is to write \(\|f\|_2\le\|f\|_1\) by “reversing” the finite-measure embedding. For the function \(f(x)=x^{-3/4}\) on \((0,1)\), this would claim \(\|f\|_2\le4\). But \(\int_0^1|f|^2\,dm=\infty\), so \(\|f\|_2=\infty\), contradicting that conclusion. The correction is that the proved estimate for \(p=1,q=2\) is \(\|f\|_1\le\|f\|_2\) when the measure is \(1\); it applies to \(f\in L^2\), not to every \(f\in L^1\). A hypothesis and an arrow both matter.
When checking an \(L^p\) claim, first identify the measure space and whether its total measure is finite. Then write the proposed inclusion as a directed statement, such as \(L^q\subseteq L^p\), and check that its hypotheses match the embedding theorem. For density in \(L^\infty\), specify the approximating class: unrestricted measurable simple functions work, while finite-measure-support simple functions cannot approximate \(1\) on an infinite-measure space.
Check Your Understanding
For each claim, identify the hypothesis or definition that decides whether it is valid.
- On an arbitrary measure space, are unrestricted finite-valued measurable simple functions dense in \(L^\infty(\mu)\) in the essential-supremum norm? Give the approximation idea.
- On \(\mathbb{R}\) with Lebesgue measure, why can a simple function vanishing outside a finite-measure set not approximate the constant function \(1\) in \(L^\infty\) within distance \(1/2\)?
- If \(\mu(X)=M\) with \(0<M<\infty\) and \(1\le p<q<\infty\), which inclusion is guaranteed, and what is its norm bound?
- For \(f(x)=x^{-3/4}\) on \((0,1)\), compute whether \(f\) belongs to \(L^1(m)\) and \(L^2(m)\).
- What two functions in the worked example on \(\mathbb{R}\) disprove the two possible \(L^1\)-\(L^2\) inclusions?