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Indicator Functions and Lp Membership

Learn how the measure of a set determines the \(L^p\) size of its indicator, and when an indicator or its scalar multiple belongs to \(L^p\).

Advanced 10 min read

What You'll Learn

  • Derive the \(L^p\) formula for an indicator from the integral of a measurable set.
  • Handle zero scalars separately when the set has infinite measure.
  • Characterize \(L^p\) membership of indicators and their scalar multiples.
  • Compute indicator norms for Lebesgue measure and counting measure.
  • Relate equality almost everywhere of indicators to null symmetric differences.
  • Identify and correct common errors involving infinite measure.

Why Indicators Reveal Set Size

In \(L^p\), the size of a function is measured by integrating its \(p\)-th power. An indicator function takes only the values zero and one, so its \(p\)-th power is again an indicator. Its integral therefore records exactly the measure of the set where it is one. This simple calculation gives a direct test for whether an indicator belongs to \(L^p\), and it also highlights a subtle issue: multiplying an indicator of an infinite-measure set by zero gives the zero function, not an expression involving the undefined product \(0\cdot\infty\).

Throughout, let \((X,\mathcal{M},\mu)\) be an arbitrary measure space; no assumption of finite measure, completeness, or \(\sigma\)-finiteness is needed. Fix a finite exponent \(p\) with \(1\le p<\infty\). Functions may be real-valued or complex-valued and are assumed measurable when \(L^p\) membership is discussed. For \(E\in\mathcal{M}\), the indicator \(\chi_E\) is the real-valued function equal to one on \(E\) and zero on \(E^c\). The integral of a nonnegative measurable function is allowed to be \(+\infty\), and \(\int\chi_E\,d\mu=\mu(E)\), including when \(\mu(E)=+\infty\).

Definition: A measurable real- or complex-valued function \(f\) belongs to \(L^p(\mu)\) if \(\int_X |f|^p\,d\mu<\infty\). Its \(L^p\) norm is \(\|f\|_p=(\int_X|f|^p\,d\mu)^{1/p}\). The space identifies functions that agree \(\mu\)-almost everywhere. For a measurable \(f\) that may not belong to \(L^p(\mu)\), we will refer to \((\int_X|f|^p\,d\mu)^{1/p}\) as its extended \(L^p\) size, allowing the value \(+\infty\); it is a norm only when \(f\in L^p(\mu)\).

The Indicator Formula and Membership Test

Theorem: Let \(E\in\mathcal{M}\) in the measure space \((X,\mathcal{M},\mu)\), and let \(1\le p<\infty\). Then \(\|\chi_E\|_p=\mu(E)^{1/p}\) when \(\mu(E)<\infty\); if \(\mu(E)=+\infty\), its extended \(L^p\) size is \(+\infty\) and \(\chi_E\notin L^p(\mu)\). More generally, for a real or complex scalar \(c\), if \(c=0\), then \(c\chi_E=0\), its integral of the \(p\)-th power is zero, and its norm is zero. If \(c\ne0\), then \(\int_X|c\chi_E|^p\,d\mu=|c|^p\mu(E)\), and its extended \(L^p\) size is \(|c|\mu(E)^{1/p}\), interpreted as \(+\infty\) if \(\mu(E)=+\infty\). Consequently, \(c\chi_E\in L^p(\mu)\) exactly when \(c=0\) or \(\mu(E)<\infty\).

Proof. At every \(x\in X\), \(|\chi_E(x)|^p=\chi_E(x)\), because \(\chi_E(x)\) is either zero or one. Thus \(\int_X|\chi_E|^p\,d\mu=\int_X\chi_E\,d\mu=\mu(E)\). If this measure is finite, taking the \(p\)-th root gives the norm; if it is infinite, the defining integral is not finite, so \(\chi_E\notin L^p(\mu)\).

Now consider \(c\chi_E\). If \(c=0\), then the function is zero at every point, even if \(\mu(E)=+\infty\). Its \(p\)-th power has integral zero, and its norm is zero. If \(c\ne0\), then pointwise \(|c\chi_E|^p=|c|^p\chi_E\). The constant \(|c|^p\) is positive and finite, so integration gives \(|c|^p\mu(E)\), with value \(+\infty\) when \(\mu(E)=+\infty\). Taking the \(p\)-th root gives the stated extended size. It is finite exactly when \(\mu(E)<\infty\). Including the separate case \(c=0\) proves the membership criterion. \(\square\)

The theorem tests membership directly from the set, without requiring any approximation of the indicator by other functions. For a nonzero scalar, the value of \(p\) changes the numerical size but not the membership decision: the integral is finite precisely when \(E\) has finite measure. If \(E\) is null, the norm is zero for every finite \(p\), although \(\chi_E\) need not be pointwise zero. This is consistent with \(L^p(\mu)\) identifying functions equal almost everywhere.

Worked Example: A Complex Scalar on a Bounded Interval

Take \(X=\mathbb{R}\) with its Lebesgue \(\sigma\)-algebra and Lebesgue measure \(m\). Let \(E=(-2,1]\), \(c=3+4i\), and \(p=2\). The interval has length \(1-(-2)=3\), so \(m(E)=3\). Also, \(|c|=\sqrt{3^2+4^2}=\sqrt{9+16}=5\). Since \(c\ne0\), the theorem gives

$$ \int_{\mathbb{R}}|c\chi_E|^2\,dm =|c|^2m(E) =5^2\cdot3 =25\cdot3 =75. $$

Therefore \(\|c\chi_E\|_2=\sqrt{75}=\sqrt{25\cdot3}=5\sqrt{3}\). Check: the function has magnitude \(5\) on a set of measure \(3\) and magnitude \(0\) elsewhere, so integrating its squared magnitude gives \(5^2\cdot3=75\), whose square root is again \(5\sqrt{3}\).

Worked Example: An Indicator on an Infinite Counting Space

Let \(X=\mathbb{N}=\{1,2,\ldots\}\), let \(\mathcal{M}\) be the collection of all subsets of \(\mathbb{N}\), and let \(\mu\) be counting measure, so \(\mu(E)\) is the number of elements of \(E\), or \(+\infty\) if \(E\) is infinite. For \(E=\mathbb{N}\) and \(p=2\), every summand \(|\chi_E(n)|^2\) equals \(1\). Thus

$$ \int_{\mathbb{N}}|\chi_{\mathbb{N}}|^2\,d\mu =\sum_{n=1}^{\infty}1 =+\infty. $$

Hence \(\chi_{\mathbb{N}}\notin L^2(\mu)\), and its extended \(L^2\) size is \(+\infty\). For a finite check, let \(E_N=\{1,\ldots,N\}\), where \(N\) is a positive integer. It has exactly \(N\) elements, each contributing \(1\), so \(\int|\chi_{E_N}|^2\,d\mu=N\) and \(\|\chi_{E_N}\|_2=\sqrt{N}\). At \(N=9\), the sum has nine terms equal to one, giving integral \(9\) and norm \(3\). As \(N\) increases without bound, these finite-set norms increase without bound, in agreement with the infinite-set calculation.

Check: direct summation over \(\mathbb{N}\) gives an infinite sum of ones, while the set formula gives \(\mu(\mathbb{N})^{1/2}=+\infty\). Both methods give the same result.

Worked Example: A Finite-Measure Set Made of Infinitely Many Intervals

On \(\mathbb{R}\) with Lebesgue measure \(m\), define \(I_n=(3n,3n+2^{-n})\) for integers \(n\ge1\), and set \(E=\bigcup_{n=1}^{\infty}I_n\). These intervals are pairwise disjoint: for each \(n\), \(3n+2^{-n}\le3n+1<3(n+1)\). Their lengths are \(2^{-n}\). By countable additivity and the geometric series formula,

$$ m(E)=\sum_{n=1}^{\infty}2^{-n} =\frac{1/2}{1-1/2} =1. $$

Thus for \(p=4\), \(\chi_E\in L^4(m)\) and \(\|\chi_E\|_4=1^{1/4}=1\). Check: \(\chi_E^4=\chi_E\), so its integral is \(m(E)=1\), and taking the fourth root again gives \(1\). Although \(E\) has infinitely many component intervals and is unbounded, its finite measure is what determines membership and the norm in this example.

Almost-Everywhere Equality of Indicators

Proposition: For \(E,F\in\mathcal{M}\) in any measure space \((X,\mathcal{M},\mu)\), \(\chi_E=\chi_F\) \(\mu\)-almost everywhere if and only if \(\mu(E\triangle F)=0\). In particular, if \(\mu(E\triangle F)=0\), then the indicators represent the same element of \(L^p(\mu)\) whenever either one belongs to \(L^p(\mu)\), and their norms are equal.

Proof. At a point \(x\), the values \(\chi_E(x)\) and \(\chi_F(x)\) differ exactly when \(x\) belongs to one of \(E,F\) but not the other. That set is \(E\triangle F=(E\setminus F)\cup(F\setminus E)\). Therefore the set on which the indicators differ is precisely \(E\triangle F\). They agree almost everywhere exactly when this set has measure zero.

If \(\mu(E\triangle F)=0\), then \(E\setminus F\) and \(F\setminus E\) are null. The sets \(E\cap F\), \(E\setminus F\) are disjoint and have union \(E\), so countable additivity gives \(\mu(E)=\mu(E\cap F)+0=\mu(E\cap F)\), including when this value is infinite. The same argument gives \(\mu(F)=\mu(E\cap F)\). Hence \(\mu(E)=\mu(F)\), and the indicator formula gives equal extended sizes and equal norms whenever finite. \(\square\)

For example, changing a measurable set by adding or removing a null set cannot change the \(L^p\) element represented by its indicator. The result concerns equality almost everywhere, not necessarily equality at every point.

A Common Pitfall: The Zero Scalar

A tempting but invalid calculation is to apply \(\int|c\chi_E|^p\,d\mu=|c|^p\mu(E)\) without separating \(c=0\) from \(0<\mu(E)=+\infty\). For example, on \(\mathbb{R}\) with Lebesgue measure, taking \(E=\mathbb{R}\), \(c=0\), and \(p=2\) would produce the expression \(0^2m(\mathbb{R})=0\cdot\infty\). That product is undefined, so it cannot establish a value for the integral.

The correction is pointwise and direct: \(0\chi_{\mathbb{R}}\) is the zero function everywhere. Therefore \(\int_{\mathbb{R}}|0\chi_{\mathbb{R}}|^2\,dm=\int_{\mathbb{R}}0\,dm=0\), and its \(L^2\) norm is zero. The product formula is used for \(c\ne0\); for \(c=0\), compute the zero function itself. Conversely, for \(c=1\) and \(E=\mathbb{R}\), the integral is \(m(\mathbb{R})=+\infty\), so \(\chi_{\mathbb{R}}\notin L^2(m)\). These two cases show why the scalar and the measure of the set must both be checked.

Check Your Understanding

In each question, state whether the function belongs to the specified \(L^p\) space and give its norm when it does.

  1. On \(\mathbb{R}\) with Lebesgue measure, let \(E=(0,5]\). Compute \(\|\chi_E\|_3\).
  2. On \(\mathbb{N}\) with counting measure, does \(2\chi_{\{2,4,6\}}\) belong to \(L^1\)? Compute its norm.
  3. On any measure space, what are the integral and norm of \(0\chi_E\) when \(\mu(E)=+\infty\)? Explain why the product formula should not be used directly.
  4. For a measurable set \(E\) and \(c\ne0\), give the necessary and sufficient condition for \(c\chi_E\in L^p(\mu)\), where \(1\le p<\infty\).
  5. Show that if \(E,F\) are measurable and \(\mu(E\triangle F)=0\), then their indicators have the same \(L^p\) norm whenever they belong to \(L^p(\mu)\).

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