Slicing Measurable Regions with Indicator Functions

Why Slice a Measurable Region?

A region in a product space can be difficult to measure directly. Sometimes, however, its slices are simple: fixing one coordinate leaves an interval, a rectangle, or another set whose measure is easy to compute. Integrating those slice measures recovers the measure of the whole region.

The underlying idea is to represent the region by its indicator function. That function is nonnegative and measurable, so Tonelli’s theorem applies under its stated hypotheses. The result is a precise formula for measuring a region by vertical or horizontal sections.

Core idea: The indicator of a region is one on the region and zero outside it. Integrating that indicator over the product space gives the region’s product measure; Tonelli rewrites this integral as an iterated integral of section indicators.

Sections and Indicator Functions

Let \((X,\mathcal A,\mu)\) and \((Y,\mathcal B,\nu)\) be measure spaces, and let \(E\in\mathcal A\otimes\mathcal B\). For \(x\in X\), the section of \(E\) at \(x\) is the subset of \(Y\) consisting of points paired with \(x\) that belong to \(E\). For \(y\in Y\), the section at \(y\) is defined in the other direction.

E_x=\{y\in Y:(x,y)\in E\}, \qquad E^y=\{x\in X:(x,y)\in E\}.

The set \(E_x\) is often called a vertical section when \(X\) is the horizontal coordinate and \(Y\) the vertical coordinate. The set \(E^y\) is then a horizontal section. The names describe the familiar Euclidean picture; the definitions themselves apply to general product spaces.

The indicator function of \(E\) is the function \(\mathbf 1_E:X\times Y\to\{0,1\}\) defined by \(\mathbf 1_E(x,y)=1\) when \((x,y)\in E\), and \(\mathbf 1_E(x,y)=0\) otherwise. Its defining relation to the sections is immediate:

\mathbf 1_E(x,y)=\mathbf 1_{E_x}(y)=\mathbf 1_{E^y}(x).

Because \(E\) belongs to the product sigma-algebra, its indicator is product-measurable. Also, its sections are measurable. The section theorem for measurable sets and Tonelli’s theorem give the measurability of the functions of section measures that appear below.

The Section Formula

We will use Tonelli’s theorem in its nonnegative form: if both measure spaces are sigma-finite and \(f:X\times Y\to[0,\infty]\) is product-measurable, then the section integrals are measurable functions and the two iterated integrals equal the product integral. The equalities hold as extended nonnegative values, so the common value may be infinite.

Theorem: Suppose \((X,\mathcal A,\mu)\) and \((Y,\mathcal B,\nu)\) are sigma-finite measure spaces, and \(E\in\mathcal A\otimes\mathcal B\). Then

(\mu\times\nu)(E) = \int_X \nu(E_x)\,\mu(dx) = \int_Y \mu(E^y)\,\nu(dy).

Proof. The function \(\mathbf 1_E\) is nonnegative and \(\mathcal A\otimes\mathcal B\)-measurable. Thus Tonelli’s theorem applies. For each \(x\in X\), its section in the \(Y\) coordinate is \(\mathbf 1_{E_x}\), and therefore

\int_Y \mathbf 1_E(x,y)\,\nu(dy) = \int_Y \mathbf 1_{E_x}(y)\,\nu(dy) = \nu(E_x).

Likewise, for each \(y\in Y\), integration in the \(X\) coordinate gives

\int_X \mathbf 1_E(x,y)\,\mu(dx) = \int_X \mathbf 1_{E^y}(x)\,\mu(dx) = \mu(E^y).

Finally, the integral of an indicator is the measure of the set it indicates:

\int_{X\times Y}\mathbf 1_E\,d(\mu\times\nu) = (\mu\times\nu)(E).

Applying Tonelli to \(\mathbf 1_E\) in either order proves both equalities in the claimed formula. \(\square\)

The sigma-finiteness conditions are part of this theorem’s hypotheses. In particular, an earlier section-integral result that assumes sigma-finiteness cannot be invoked for an arbitrary measure space. Here, both \(\mu\) and \(\nu\) are assumed sigma-finite before the theorem is applied.

Worked Examples

Worked Example: Measuring a Rectangle by Its Sections

Take \(X=Y=\mathbb R\) with Lebesgue measure \(m\), and let \(E=[0,2]\times[1,4]\). The vertical section is \([1,4]\) for \(x\in[0,2]\), and it is empty for \(x\notin[0,2]\). Its length is therefore \(3\) on an interval of \(x\)-values of length \(2\), and zero elsewhere. The section formula gives

(m\times m)(E) = \int_{\mathbb R}m(E_x)\,dx = \int_0^2 3\,dx = 6.

Reversing the roles, \(E^y=[0,2]\) for \(y\in[1,4]\) and is empty otherwise. Hence the horizontal calculation is

(m\times m)(E) = \int_{\mathbb R}m(E^y)\,dy = \int_1^4 2\,dy = 6.

Both computations use the same product-measurable rectangle and give the same product measure.

Worked Example: The Region Under a Curved Boundary

Consider the subset of the unit square below the curve \(y=x^2\): \(E=\{(x,y)\in[0,1]^2:0\le y\le x^2\}\). The defining inequalities involve continuous functions, so \(E\) is a Borel, and hence product-measurable, subset of \(\mathbb R^2\).

Fix \(x\in[0,1]\). The vertical section is \(E_x=[0,x^2]\), including the singleton \(\{0\}\) when \(x=0\). Its length is \(x^2\), so

(m\times m)(E) = \int_0^1 m(E_x)\,dx = \int_0^1 x^2\,dx = \frac13.

Now fix \(y\in[0,1]\). The condition \(y\le x^2\), together with \(x\ge0\), is equivalent to \(x\ge\sqrt y\). Thus \(E^y=[\sqrt y,1]\), whose length is \(1-\sqrt y\). Slicing horizontally gives

(m\times m)(E) = \int_0^1 m(E^y)\,dy = \int_0^1(1-\sqrt y)\,dy = 1-\frac23 = \frac13.

The endpoint sections also agree with these length formulas: at \(y=0\) the section is all of \([0,1]\), of length \(1\), and at \(y=1\) it is a singleton, of length \(0\).

Worked Example: A Slanted Band in a Rectangle

Let \(E=\{(x,y)\in\mathbb R^2:0\le x\le2,\ x\le y\le x+1\}\). This is a Borel set because it is defined by inequalities involving continuous functions. For each \(x\in[0,2]\), the vertical section is the interval \(E_x=[x,x+1]\), of length \(1\). Outside that \(x\)-interval the section is empty. Therefore

(m\times m)(E) = \int_{\mathbb R}m(E_x)\,dx = \int_0^2 1\,dx = 2.

To check this using horizontal sections, the inequalities for a fixed \(y\) require \(0\le x\le2\), \(x\le y\), and \(x\ge y-1\). For \(0\le y\le1\), these reduce to \(0\le x\le y\), so the section length is \(y\). For \(1\le y\le2\), the section is \([y-1,y]\), of length \(1\). For \(2\le y\le3\), it is \([y-1,2]\), of length \(3-y\). For \(y\) outside \([0,3]\), it is empty. Thus

\begin{aligned} (m\times m)(E) &=\int_0^1 y\,dy+\int_1^2 1\,dy+\int_2^3(3-y)\,dy\\ &=\frac12+1+\frac12\\ &=2. \end{aligned}

The breakpoints \(0,1,2,3\) come from the changes in which of the bounds on \(x\) is active. Writing those bounds explicitly is a reliable way to avoid missing part of a section.

A Useful Consequence: Bounding a Region

Section measures can give a quick bound even when their exact integral is not available. The following consequence makes the required conditions explicit.

Proposition: Suppose \(\mu\) and \(\nu\) are sigma-finite, \(E\in\mathcal A\otimes\mathcal B\), and \(A_0\in\mathcal A\) has finite measure. If \(\nu(E_x)=0\) for \(x\notin A_0\) and \(\nu(E_x)\le C\) for \(x\in A_0\), where \(C\) is a finite nonnegative constant, then

(\mu\times\nu)(E)\le C\,\mu(A_0).

Proof. By the section formula,

(\mu\times\nu)(E) = \int_X\nu(E_x)\,\mu(dx) = \int_{A_0}\nu(E_x)\,\mu(dx).

The integrand is nonnegative and at most \(C\) on \(A_0\). Monotonicity of the integral therefore gives

\int_{A_0}\nu(E_x)\,\mu(dx) \le \int_{A_0}C\,\mu(dx) = C\,\mu(A_0).

Since \(\mu(A_0)<\infty\) and \(C<\infty\), this is a finite upper bound. \(\square\)

For instance, if every vertical section of a region inside a base set of finite measure has length at most \(C\), then the region’s area is at most \(C\) times the measure of that base set. This estimate depends on the section bound holding for every \(x\) in the base, or at least almost everywhere with respect to \(\mu\); changing section values on a \(\mu\)-null set does not change the integral.

Interpreting the Formula Carefully

The section formula integrates a nonnegative function, \(x\mapsto\nu(E_x)\), or \(y\mapsto\mu(E^y)\). Its value is allowed to be \(+\infty\). If one iterated integral is finite, then the product measure is finite and the other iterated integral has that same finite value. No cancellation is involved.

This is why the indicator method is especially safe for measuring regions. It uses Tonelli’s theorem for a nonnegative integrand, rather than relying on a change in integration order for a signed or complex function. Those settings require their own hypotheses and results.

A second point is that the theorem is stated for \(E\in\mathcal A\otimes\mathcal B\). This ensures that the indicator is measurable on the product space. If a set is described only as measurable in a completion of the product measure, the product-sigma-algebra statement above does not by itself establish all the required section properties. One should check the relevant completed-measure version before applying a section formula.

Finally, choosing a slicing direction is a computational decision, not a change in the region. One direction may produce a constant section length; another may require a piecewise formula. Both are valid when the hypotheses hold, and both must return the same product measure.

Check Your Understanding

Use the definitions and section formula to answer the following questions.

For \(E\in\mathcal A\otimes\mathcal B\), write the vertical section \(E_x\) and horizontal section \(E^y\) in set-builder notation.
Which theorem justifies integrating the indicator of \(E\) in either order, and what measure-space assumption is needed here?
For the region \(E=\{(x,y)\in[0,1]^2:0\le y\le x^2\}\), what is the length of the horizontal section at a fixed \(y\in[0,1]\)?
In the slanted-band example, why does the horizontal section formula change at \(y=1\) and \(y=2\)?
Why is the indicator-function approach a nonnegative integration problem even when the region has infinite measure?