From Simple Values to an \(L^p\) Norm
A simple function takes only finitely many values, so its \(L^p\) norm reduces to a finite computation: raise the absolute values of those values to the \(p\)th power, weight them by the measures of the sets where they occur, add, and take a \(p\)th root. The essential bookkeeping is to use disjoint sets on which the function is actually constant. An expression involving overlapping indicator functions may not yet be in that form.
Let \((X,\mathcal{M},\mu)\) be a measure space; no assumption of finiteness, sigma-finiteness, or completeness is imposed. The functions considered here are measurable, finite-valued real- or complex-valued simple functions. Fix \(p\) with \(1\le p<\infty\). For a measurable function \(f\), its \(L^p\) norm, when finite, is \(\|f\|_p=(\int_X |f|^p\,d\mu)^{1/p}\). The space \(L^p(\mu)\) consists of the almost-everywhere equivalence classes of measurable functions for which this integral is finite. The \(L^\infty\) norm is not part of this computation.
When \(c_j\ne0\) and \(\mu(E_j)=\infty\), the corresponding term is \(+\infty\), so the extended integral is infinite. This does not define a finite \(L^p\) norm. The formula also applies when the sets do not cover \(X\): the function is zero on the uncovered part.
The Finite-Sum Formula
Proof. Since the sets \(E_j\) are pairwise disjoint, at each \(x\in E_j\) the function \(f(x)\) equals \(c_j\), and outside their union it equals zero. Therefore \(|f|^p=\sum_{j=1}^N |c_j|^p\chi_{E_j}\) pointwise. The integral of a nonnegative simple function is the sum of each nonnegative coefficient times the measure of its measurable set. Applying this definition gives \(\int_X|f|^p\,d\mu=\sum_{j=1}^N|c_j|^p\mu(E_j)\), with zero terms omitted to avoid the undefined product \(0\cdot\infty\). By the definition of \(L^p(\mu)\), \(f\) belongs to that space exactly when the integral is finite; taking its \(p\)th root gives the stated norm. \(\square\)
The coefficients may be negative or complex, but the formula uses their absolute values. For example, a value \(-4\) contributes \(4^p\), not \((-4)^p\). The measure weights also matter: a value occurring on a set of measure \(5\) contributes five times as much as the same value on a set of measure \(1\).
Proof. If \(c_j=0\), both contributions are zero under the convention in the theorem. If \(c_j\ne0\), finite additivity of the measure on the disjoint measurable sets gives \(\mu(E_j)=\sum_{k=1}^r\mu(E_{jk})\). This equality holds for nonnegative extended values as well: if \(\mu(E_j)=\infty\), at least one of the finitely many pieces has infinite measure, since a finite sum of finite numbers is finite. Multiplication by the finite positive number \(|c_j|^p\) therefore gives the same extended contribution before and after refinement. Applying this to each cell proves the assertion. \(\square\)
This proposition permits us to turn a formula involving overlapping sets into the disjoint level sets needed for the norm calculation. It also shows that further subdividing a constant-value region cannot alter the answer.
Worked Norm Computations
Worked Example: Counting Measure on a Four-Point Set
Let \(X=\{a,b,c,d\}\), let \(\mathcal{M}\) be the set of all subsets of \(X\), and let \(\mu\) be counting measure. Suppose \(f(a)=f(b)=3\), \(f(c)=-2\), and \(f(d)=1\). Compute \(\|f\|_2\).
The disjoint value sets are \(E_1=\{a,b\}\), \(E_2=\{c\}\), and \(E_3=\{d\}\). Their measures are \(\mu(E_1)=2\), \(\mu(E_2)=1\), and \(\mu(E_3)=1\). The finite-sum formula gives \[ \|f\|_2^2=|3|^2\cdot2+|-2|^2\cdot1+|1|^2\cdot1=9\cdot2+4\cdot1+1\cdot1=23. \] Thus \(\|f\|_2=\sqrt{23}\). Check: counting measure sums \(|f|^2\) over the four points, giving \(9+9+4+1=23\), hence again \(\|f\|_2=\sqrt{23}\).
Worked Example: Different Values on Intervals
On \((\mathbb{R},\mathcal{L},m)\), let \(f=2\) on \(E_1=[0,1)\), \(f=-1\) on \(E_2=[1,3)\), and \(f=3\) on \(E_3=[3,4]\), with \(f=0\) elsewhere. Compute \(\|f\|_3\). Here \(m\) is Lebesgue measure, and these three measurable sets are disjoint.
The lengths are \(m(E_1)=1-0=1\), \(m(E_2)=3-1=2\), and \(m(E_3)=4-3=1\). Therefore \[ \|f\|_3^3=|2|^3\cdot1+|-1|^3\cdot2+|3|^3\cdot1 =8\cdot1+1\cdot2+27\cdot1=37, \] so \(\|f\|_3=\sqrt[3]{37}\). Check: integrating \(|f|^3\) separately over the same intervals gives \(8(1)+1(2)+27(1)=37\); taking the cube root agrees with the formula.
Worked Example: Refining Overlapping Indicator Sets
On \(([0,4),\mathcal{L}|_{[0,4)},m)\), define \[ f=2\chi_{[0,2)}+3\chi_{[1,3)}. \] Compute \(\|f\|_2\). The two sets in this expression overlap, so their coefficients cannot be treated as values on disjoint regions.
On \([0,1)\), only the first indicator is \(1\), so \(f=2\). On \([1,2)\), both are \(1\), so \(f=2+3=5\). On \([2,3)\), only the second is \(1\), so \(f=3\). On \([3,4)\), both are zero. These four intervals are disjoint and have measures \(1,1,1,1\), respectively. Thus \[ \|f\|_2^2=|2|^2\cdot1+|5|^2\cdot1+|3|^2\cdot1 =4+25+9=38, \] and \(\|f\|_2=\sqrt{38}\). Check: \(|f|^2\) has the constant values \(4,25,9,0\) on four intervals of length \(1\); its integral is \(4+25+9+0=38\), giving the same norm.
Worked Example: A Nonzero Value on an Infinite-Measure Set
On \((\mathbb{R},\mathcal{L},m)\), let \(f=\chi_{(0,\infty)}\). For any fixed \(1\le p<\infty\), determine whether \(f\in L^p(m)\).
The function has value \(1\) on \((0,\infty)\), whose Lebesgue measure is infinite, and value zero elsewhere. The formula yields \[ \int_{\mathbb{R}}|f|^p\,dm=|1|^p m((0,\infty))=1\cdot\infty=\infty. \] Therefore \(f\notin L^p(m)\), and it has no finite \(L^p\) norm. Check: \(|f|^p=1\) throughout the positive half-line, so its integral over \((0,R)\) is \(R\) for every \(R>0\); these integrals are unbounded as \(R\) increases. In contrast, the zero function on all of \(\mathbb{R}\) has integral zero, since its only coefficient is zero, despite the underlying space having infinite measure.
A Common Partitioning Error
A frequent mistake is to apply the sum formula directly to overlapping indicator sets as though each coefficient described a separate, disjoint region. In the third example, the incorrect calculation \[ \|f\|_2^2\overset{\text{wrong}}{=} |2|^2m([0,2))+|3|^2m([1,3))=4\cdot2+9\cdot2=26 \] counts the overlap twice without accounting for the fact that the function has value \(5\) there. The correct partition has a region of measure \(1\) where the value is \(2\), a region of measure \(1\) where the value is \(5\), and a region of measure \(1\) where the value is \(3\). Its contribution is \(4+25+9=38\), not \(26\). The repair is to determine the actual values of \(f\) on disjoint measurable pieces before using the formula.
For a finite-valued simple function, the practical procedure is therefore: collect the sets on which each value occurs; make the resulting value sets disjoint; compute the measure of each set; use \(|c_j|^p\), not \(c_j^p\); and check whether any nonzero value occurs on a set of infinite measure. This last check distinguishes a finite \(L^p\) norm from an infinite integral.
Check Your Understanding
For each computation, first identify disjoint value sets and then apply the finite-sum formula.
- On a three-point set with counting measure, \(f\) takes values \(2,-1,2\). Compute \(\|f\|_2\).
- On \([0,3)\) with Lebesgue measure, \(f\) equals \(1\) on \([0,2)\) and \(-2\) on \([2,3)\). Compute \(\|f\|_1\) and \(\|f\|_2\).
- For \(f=2\chi_A+3\chi_B\), where \(A\) and \(B\) are disjoint measurable sets of finite measures \(2\) and \(1\), compute \(\|f\|_2\).
- Explain why the formula cannot be applied to overlapping indicator sets by simply adding the separate coefficient contributions.
- On a measure space with \(\mu(E)=\infty\), what is \(\int |c\chi_E|^p\,d\mu\) when \(c\ne0\)? What is it when \(c=0\)?