Integrating Over a Measurable Subset

Restricting an Integral to a Measurable Set

An integral over a measurable subset records only the contribution from that part of the space. The standard way to express this is to multiply the function by the indicator of the subset: the product agrees with the function on the subset and is zero elsewhere. For simple functions, this operation has a direct computational interpretation: intersect each level set with the subset before measuring it.

Throughout, \((X,\mathcal{M},\mu)\) is an arbitrary measure space; no assumption of finiteness, sigma-finiteness, or completeness is needed. Functions in this tutorial are measurable and take values in \([0,\infty]\). We use the nonnegative Lebesgue integral, defined as the supremum of the integrals of nonnegative simple functions below the function. A nonnegative simple function has finitely many finite values and measurable level sets. Its integral is the sum of each value times the measure of the set where that value occurs.

The function \(f\chi_A\) is measurable and nonnegative, so its integral is defined on every measure space, with \(+\infty\) allowed. The subset integral is not an average: it is not divided by \(\mu(A)\). In particular, its definition makes sense whether or not \(A\) has finite measure.

Simple Functions: Restrict the Level Sets

Suppose \(\varphi\) is a nonnegative simple function and \(A\in\mathcal{M}\). If \(\varphi=\sum_{j=1}^r c_j\chi_{E_j}\), where the \(E_j\) are disjoint measurable sets and \(c_j\ge0\), then restricting to \(A\) replaces each \(E_j\) by \(E_j\cap A\). Equivalently, one can use the superlevel sets of \(\varphi\).

Proof. For each \(x\in X\), if \(\varphi(x)=0\), then every indicator \(\chi_{\{\varphi\ge a_j\}}(x)\) is zero. If \(\varphi(x)=a_\ell\) for some \(1\le\ell\le k\), then \(\varphi(x)\ge a_j\) holds exactly when \(j\le\ell\). Consequently, \(\sum_{j=1}^k(a_j-a_{j-1})\chi_{\{\varphi\ge a_j\}}(x)=\sum_{j=1}^{\ell}(a_j-a_{j-1})=a_\ell=\varphi(x)\). This proves the pointwise identity \(\varphi=\sum_{j=1}^k(a_j-a_{j-1})\chi_{\{\varphi\ge a_j\}}\). Multiplying by \(\chi_A\) replaces each superlevel set by its intersection with \(A\). The integral of a nonnegative simple function is the sum of its coefficients times the measures of its disjoint level sets; applying that definition to this finite decomposition gives the formula. If there are no positive values, then \(\varphi=0\) everywhere and both sides are zero. \(\square\)

The level sets in this formula are nested: \(\{\varphi\ge a_{j+1}\}\subseteq\{\varphi\ge a_j\}\). Their intersections with \(A\) are nested as well. The successive differences \(a_j-a_{j-1}\) account for the value of \(\varphi\) accumulated at each level. The intersection is essential: measuring the full superlevel set instead would compute a contribution from points outside \(A\).

The Subset Integral as a Supremum

Proof. By definition, \(\int_X f\chi_A\,d\mu\) is the supremum of \(\int_X\psi\,d\mu\) over nonnegative simple \(\psi\le f\chi_A\). Every such \(\psi\) vanishes on \(A^c\), since there \(0\le\psi\le f\chi_A=0\). Hence \(\psi=\psi\chi_A\) pointwise. Also \(\psi\le f\) on \(A\), and \(\psi=0\le f\) on \(A^c\), so \(\psi\) is one of the simple functions appearing on the right-hand side of the theorem. Its integral over \(X\) equals \(\int_A\psi\,d\mu\) by the definition of subset integral.

Conversely, if \(\varphi\) is nonnegative simple and \(\varphi\le f\), then \(\varphi\chi_A\) is nonnegative simple and \(\varphi\chi_A\le f\chi_A\) pointwise. Its integral over \(X\) is \(\int_A\varphi\,d\mu\). Thus each value used in the supremum on the right also occurs among the simple minorants defining \(\int_X f\chi_A\,d\mu\), and every value in the latter supremum occurs in the former. The two suprema are equal, including when they are infinite. \(\square\)

For a simple \(f\), the theorem and the level-set formula give a finite computation. For a general nonnegative measurable \(f\), the theorem says that restricting the integral is compatible with approximation from below: one may integrate simple minorants only on \(A\). It does not assert that a particular simple minorant reaches the supremum.

Disjoint Subsets and a Common Error

Proof. Since \(A\) and \(B\) are disjoint, at each point \(x\) at most one of \(\chi_A(x)\) and \(\chi_B(x)\) is \(1\), and \(\chi_{A\cup B}=\chi_A+\chi_B\). Under the stated convention for products, this gives the pointwise identity \(f\chi_{A\cup B}=f\chi_A+f\chi_B\), also when \(f(x)=\infty\). The additivity of the nonnegative integral for arbitrary nonnegative measurable functions now gives \(\int_X f\chi_{A\cup B}\,d\mu=\int_X f\chi_A\,d\mu+\int_X f\chi_B\,d\mu\). Applying the definition of each subset integral proves the result. \(\square\)

A common mistake is to use the measures of the whole-space level sets in the level-set formula, even though the desired integral is over \(A\). In the first worked example, doing that would give \(1\cdot4+4\cdot2=12\): the whole level set \(\{\varphi\ge1\}\) has length \(4\), and \(\{\varphi\ge5\}\) has length \(2\). This is \(\int_X\varphi\,dm\), not \(\int_A\varphi\,dm\). The correction is to intersect first: the relevant measures are \(2\) and \(3/2\), giving \(1\cdot2+4\cdot(3/2)=8\). The subset determines which points contribute at every level.