Restricting an Integral
Integration over a measurable set is not a different kind of integral: it can be written as an integral over the whole space after multiplying the integrand by an indicator. This simple representation makes restriction compatible with linearity, absolute-value bounds, and decomposition of the domain. It also works on measure spaces of infinite measure, provided the function being integrated is integrable.
Throughout, \((X,\mathcal{M},\mu)\) is an arbitrary measure space; it need not be finite or sigma-finite. Functions in this tutorial are real-valued and measurable. A function \(f\) is integrable if \(\int_X |f|\,d\mu<\infty\). For a measurable set \(E\in\mathcal{M}\), its indicator is \(\chi_E(x)=1\) when \(x\in E\), and \(\chi_E(x)=0\) when \(x\notin E\). We use linearity, monotonicity, and the integral triangle inequality for integrable functions established in the earlier material on the Lebesgue integral.
The notation \(\int_E f\,d\mu\) refers to restricting the domain of integration, not to changing the measure space or multiplying the answer by the measure of \(E\). The indicator formula is pointwise: \(f\chi_E\) agrees with \(f\) on \(E\) and is zero on \(E^c\). If two functions agree \(\mu\)-almost everywhere, their restricted integrals over every measurable \(E\) agree as well, since their products with \(\chi_E\) still agree almost everywhere.
Restriction Bounds
Proof. The product \(f\chi_E\) is measurable because \(f\) and \(\chi_E\) are measurable. Pointwise, \(|f\chi_E|=|f|\chi_E\le |f|\), and monotonicity of the nonnegative integral gives \(\int_X|f\chi_E|\,d\mu\le\int_X|f|\,d\mu<\infty\). Thus \(f\chi_E\) is integrable. Apply the integral triangle inequality to \(f\chi_E\), then use the definition of the restricted integral: $$ \left|\int_E f\,d\mu\right| =\left|\int_X f\chi_E\,d\mu\right| \le\int_X|f\chi_E|\,d\mu =\int_E|f|\,d\mu. $$ Finally, \(|f|\chi_E\le |f|\) implies \(\int_E|f|\,d\mu\le\int_X|f|\,d\mu\), again by monotonicity. \(\square\)
The first bound is local to \(E\); it can be much sharper than the bound using the integral over all of \(X\). It also accounts for cancellation: the signed integral may be small even when the integral of the absolute value is not.
Worked Example: A Restriction with Cancellation
Let \(X=[0,4]\), let \(\mathcal{M}\) be the Lebesgue-measurable subsets of \(X\), and let \(\mu\) be Lebesgue measure \(m\) on \(X\). Set \(f(x)=x-1\) and \(E=[0,2]\). Compute the restricted integral and the absolute-value bound.
Since \(f\) is bounded and \(X\) has finite measure, \(f\) is integrable. By the indicator definition, \(f\chi_E\) equals \(x-1\) on \([0,2]\) and zero on \((2,4]\). Therefore $$ \int_E f\,dm=\int_0^2(x-1)\,dx =\left[\frac{x^2}{2}-x\right]_0^2 =(2-2)-0=0. $$ For the bound, \(|x-1|=1-x\) on \([0,1]\) and \(|x-1|=x-1\) on \([1,2]\), so $$ \int_E|f|\,dm =\int_0^1(1-x)\,dx+\int_1^2(x-1)\,dx =\frac12+\frac12=1. $$ Thus \(\left|\int_E f\,dm\right|=0\le1=\int_E|f|\,dm\). As a check, the positive and negative contributions to \(\int_E f\,dm\) are \(\int_1^2(x-1)\,dx=1/2\) and \(\int_0^1(x-1)\,dx=-1/2\); they cancel. On the complement \((2,4]\), the integral is \(\int_2^4(x-1)\,dx=[x^2/2-x]_2^4=4\), and the whole-space integral is \(\int_0^4(x-1)\,dx=4\), consistent with \(0+4=4\).
Additivity Across a Partition
A finite measurable partition of a measurable set \(E\) is a finite collection of pairwise disjoint measurable sets whose union is \(E\). The sets may include empty sets; these contribute zero. The key pointwise identity for a finite partition \(E_1,\ldots,E_N\) is \(\chi_E=\sum_{j=1}^N\chi_{E_j}\). Multiplying by \(f\) and using linearity gives the corresponding integral formula.
Proof. Pairwise disjointness means that for each \(x\in X\), at most one of the indicators \(\chi_{E_j}(x)\) is \(1\). Their sum is consequently \(1\) exactly when \(x\in E\), and is \(0\) otherwise. Thus \(\chi_E=\sum_{j=1}^N\chi_{E_j}\) pointwise. Multiplying by \(f\) gives \(f\chi_E=\sum_{j=1}^N f\chi_{E_j}\). Each term is integrable by the restriction proposition, so linearity applies and yields $$ \int_E f\,d\mu =\int_X f\chi_E\,d\mu =\sum_{j=1}^N\int_X f\chi_{E_j}\,d\mu =\sum_{j=1}^N\int_{E_j}f\,d\mu. $$ If \(E=X\), then \(\chi_E=1\) everywhere, giving the stated special case. \(\square\)
There is also a countable version. It requires a little more care: finite linearity alone does not justify passing from partial sums to an infinite sum. The following proof uses the Monotone Convergence Theorem for nonnegative functions, which applies without a finite-measure assumption.
Proof. Set \(A_N=\bigcup_{n=1}^N E_n\). The sets \(A_N\) increase to \(E\). Since \(f\) is integrable, its positive and negative parts \(f^+=\max(f,0)\) and \(f^-=\max(-f,0)\) are nonnegative integrable functions with \(f=f^+-f^-\). The sequences \(f^+\chi_{A_N}\) and \(f^-\chi_{A_N}\) increase pointwise to \(f^+\chi_E\) and \(f^-\chi_E\), respectively. Applying the Monotone Convergence Theorem separately gives convergence of their integrals. All four limiting integrals are finite, since each is bounded by \(\int_X|f|\,d\mu\). Subtracting the two convergent sequences therefore gives \(\int_{A_N}f\,d\mu\to\int_E f\,d\mu\). By the finite partition formula, \(\int_{A_N}f\,d\mu=\sum_{n=1}^N\int_{E_n}f\,d\mu\), so the partial sums converge to \(\int_E f\,d\mu\).
To check absolute convergence, apply the Monotone Convergence Theorem to the increasing nonnegative functions \(|f|\chi_{A_N}\). It gives $$ \sum_{n=1}^{\infty}\int_{E_n}|f|\,d\mu =\lim_{N\to\infty}\int_{A_N}|f|\,d\mu =\int_E|f|\,d\mu \le\int_X|f|\,d\mu<\infty. $$ For every \(n\), the restriction bound gives \(\left|\int_{E_n}f\,d\mu\right|\le\int_{E_n}|f|\,d\mu\). Summing these inequalities proves absolute convergence of the series of restricted integrals. \(\square\)
Worked Example: Restricting a Series Under Counting Measure
Let \(X=\mathbb{N}=\{1,2,\ldots\}\), let \(\mathcal{M}\) be all subsets of \(\mathbb{N}\), and let \(\mu\) be counting measure. Define \(f(n)=(-1)^{n+1}2^{-n}\) and let \(E\) be the set of even positive integers. Compute \(\int_E f\,d\mu\) and verify the partition formula.
For counting measure, the integral of a nonnegative function is its sum over \(\mathbb{N}\); hence \(\int_{\mathbb{N}}|f|\,d\mu=\sum_{n=1}^{\infty}2^{-n}=1\), so \(f\) is integrable. The strict membership condition for \(E\) is that \(n\) is even, or \(n=2k\) for an integer \(k\ge1\). At those points \(f(2k)=-2^{-2k}=-4^{-k}\). Therefore $$ \int_E f\,d\mu =\sum_{k=1}^{\infty}-4^{-k} =-\frac{1/4}{1-1/4} =-\frac13. $$ The absolute integral on \(E\) is \(\sum_{k=1}^{\infty}4^{-k}=1/3\), so the restriction bound holds with equality.
The complement consists of odd positive integers \(n=2k-1\), \(k\ge1\). There \(f(2k-1)=2^{-(2k-1)}=2\cdot4^{-k}\), and thus $$ \int_{E^c}f\,d\mu =\sum_{k=1}^{\infty}2\cdot4^{-k} =2\cdot\frac{1/4}{1-1/4} =\frac23. $$ The whole sum is \(\sum_{n=1}^{\infty}(-1)^{n+1}2^{-n}=1/2-1/4+1/8-\cdots=1/3\), also obtained by summing the geometric series with first term \(1/2\) and ratio \(-1/2\). The partition check gives \(-1/3+2/3=1/3\), as required.
Overlapping Sets and a Common Error
Additivity in its partition form requires disjoint sets. For two possibly overlapping measurable sets \(A,B\in\mathcal{M}\), the pointwise identity \(\chi_A+\chi_B=\chi_{A\cup B}+\chi_{A\cap B}\) gives a corrected formula instead.
Proof. At a point in neither set, all four indicators in the stated identity vanish. At a point in exactly one of \(A,B\), the left side and right side of the indicator identity are both \(1\). At a point in both sets, the left side is \(2\), and the right side is \(1+1=2\). Thus the identity holds pointwise. Multiply it by \(f\) and integrate. Each restricted function is integrable, so linearity gives the formula. \(\square\)
Worked Example: Correcting for an Overlap
On \(X=[0,3]\) with Lebesgue measure, let \(f=1\) on \([0,2]\) and \(f=-2\) on \((2,3]\). Let \(A=[0,2.5]\) and \(B=[1.5,3]\). Compute the union, the overlap, and the two separate restricted integrals. Endpoints do not affect these integrals because finite sets have Lebesgue measure zero.
On \(A\), the part \([0,2]\) contributes \(1\cdot2=2\), and the part \((2,2.5]\) contributes \((-2)\cdot0.5=-1\). Hence \(\int_Af\,dm=2-1=1\). On \(B\), the part \([1.5,2]\) contributes \(1\cdot0.5=0.5\), and the part \((2,3]\) contributes \((-2)\cdot1=-2\). Hence \(\int_Bf\,dm=0.5-2=-1.5\).
The intersection consists, up to its endpoints, of \([1.5,2]\) and \((2,2.5]\), so $$ \int_{A\cap B}f\,dm=(1)(0.5)+(-2)(0.5)=0.5-1=-0.5. $$ The union is all of \([0,3]\), and its integral is \((1)(2)+(-2)(1)=0\). The overlap-corrected formula gives \(1+(-1.5)-(-0.5)=0\), matching the direct computation.
Common mistake: treating an integral as proportional to the measure of the set. For example, on \(X=[0,2]\) with Lebesgue measure, take \(f(x)=x\) and \(E=[0,1]\). A mistaken proportionality calculation would give $$ \frac{m(E)}{m(X)}\int_X f\,dm =\frac12\cdot\int_0^2x\,dx =\frac12\cdot2=1. $$ But the actual restricted integral is \(\int_E f\,dm=\int_0^1x\,dx=1/2\). The correction is to use the indicator: \(\int_E f\,dm=\int_X x\chi_{[0,1]}(x)\,dm=1/2\). A measure proportion would only encode the right answer under extra conditions on \(f\); it is not the general restriction rule.
Check Your Understanding
Use the indicator definition and state which hypotheses justify each integral or additivity step.
- For integrable real-valued \(f\) and measurable \(E\), prove \(\int_{E^c}f\,d\mu=\int_Xf\,d\mu-\int_Ef\,d\mu\).
- On \([0,1]\) with Lebesgue measure, take \(f(x)=2x-1\) and \(E=[0,1/2]\). Compute \(\int_Ef\,dm\) and \(\int_E|f|\,dm\), and check the restriction bound.
- Explain why finite additivity does not by itself justify the countable partition formula. Which convergence theorem justifies the limiting step in the proof?
- For measurable \(A,B\), use indicator functions to derive the formula for \(\int_{A\cup B}f\,d\mu\) when \(A\) and \(B\) overlap.
- Give an integrable function and measurable set for which \(\int_E f\,d\mu=0\) but \(\int_E|f|\,d\mu>0\), and identify the cancellation.