Tonelli's Identity for Nonnegative Simple Functions

From Rectangles to Simple Functions

For a rectangle \(A\times B\), the product measure is \((\mu\times\nu)(A\times B)=\mu(A)\nu(B)\). The rectangular integral identity therefore gives the same value whether one integrates its indicator on the product space or first integrates one section and then the other. A nonnegative simple function is a finite combination of indicators, but its measurable level sets need not be rectangles. The key step is to establish the indicator identity for every measurable set in the product sigma-algebra; finite linearity then gives the simple-function formula.

Let \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\) be finite measure spaces, so \(\mu(X)<\infty\) and \(\nu(Y)<\infty\). Use the product measure \(\mu\times\nu\) on \((X\times Y,\mathcal{M}\otimes\mathcal{N})\). No completeness assumption is made. For \(E\in\mathcal{M}\otimes\mathcal{N}\), write \(E_x=\{y\in Y:(x,y)\in E\}\) and \(E^y=\{x\in X:(x,y)\in E\}\). The section results established earlier ensure that these sections are measurable and that \(x\mapsto\nu(E_x)\) and \(y\mapsto\mu(E^y)\) are measurable nonnegative functions.

Why the Indicator Identity Holds

We first explain the set-level step. The proof uses two standard results established earlier: the Monotone Convergence Theorem for nonnegative measurable functions, and uniqueness of finite product measures from their values on measurable rectangles. In particular, for a nonnegative measurable sequence increasing pointwise to a limit, the integral of the limit equals the limit of the integrals.

Proof. Define \(\lambda_X(E)=\int_X\nu(E_x)\,d\mu(x)\) for \(E\in\mathcal{M}\otimes\mathcal{N}\). The section measurability result makes this integral well-defined. Also, \(\lambda_X(\varnothing)=0\). To verify countable additivity, let \(E_1,E_2,\ldots\) be pairwise disjoint members of \(\mathcal{M}\otimes\mathcal{N}\), and put \(E=\bigcup_{k=1}^\infty E_k\). For each \(x\), the sections \((E_k)_x\) are pairwise disjoint and \(E_x=\bigcup_k(E_k)_x\). Countable additivity of \(\nu\) gives \(\nu(E_x)=\sum_{k=1}^\infty\nu((E_k)_x)\). The finite partial sums are measurable, nonnegative, and increase to \(\nu(E_x)\). By the Monotone Convergence Theorem, \(\lambda_X(E)=\sum_{k=1}^\infty\lambda_X(E_k)\). Thus \(\lambda_X\) is a measure.

For a measurable rectangle \(A\times B\), the section \((A\times B)_x\) equals \(B\) when \(x\in A\) and equals \(\varnothing\) when \(x\notin A\). Hence \(\lambda_X(A\times B)=\int_X \nu(B)\chi_A(x)\,d\mu(x)=\mu(A)\nu(B)\). This is exactly \((\mu\times\nu)(A\times B)\). Both measures are finite: in particular, \(\lambda_X(X\times Y)=\mu(X)\nu(Y)<\infty\). The uniqueness theorem for finite product measures now gives \(\lambda_X(E)=(\mu\times\nu)(E)\) for every \(E\in\mathcal{M}\otimes\mathcal{N}\).

For the other order, define \(\lambda_Y(E)=\int_Y\mu(E^y)\,d\nu(y)\). The sections \(E^y\) are measurable, so the integrand is measurable. If \(E_k\) are pairwise disjoint, then for each \(y\), \(\mu((\bigcup_k E_k)^y)=\sum_k\mu(E_k^y)\). Applying the Monotone Convergence Theorem to the increasing partial sums proves countable additivity of \(\lambda_Y\). On \(A\times B\), its value is \(\mu(A)\nu(B)\), and its total mass is \(\mu(X)\nu(Y)<\infty\). Finite product-measure uniqueness gives \(\lambda_Y(E)=(\mu\times\nu)(E)\). Combining the two equalities proves the lemma. \(\square\)

Proof for Simple Functions

A nonnegative simple function here means a measurable function with finitely many finite values in \([0,\infty)\). It can be written using pairwise disjoint measurable sets as \(f=\sum_{j=1}^r a_j\chi_{E_j}\), where \(a_j\ge0\) and \(E_j\in\mathcal{M}\otimes\mathcal{N}\). A zero-valued part may be omitted. Disjointness makes the simple-function integral formula direct, and the finite measure assumptions ensure the resulting integrals are finite.

Proof of the theorem. For each \(x\in X\), the sections \((E_j)_x\) are measurable and pairwise disjoint. Thus \(f(x,\cdot)=\sum_{j=1}^r a_j\chi_{(E_j)_x}\), so the simple-function integral formula gives \(\int_Y f(x,y)\,d\nu(y)=\sum_{j=1}^r a_j\nu((E_j)_x)\). Each term is a measurable function of \(x\), so this finite sum is measurable. Integrating it and using finite additivity of the integral gives \[ \int_X\left(\int_Y f(x,y)\,d\nu(y)\right)d\mu(x) =\sum_{j=1}^r a_j\int_X\nu((E_j)_x)\,d\mu(x) =\sum_{j=1}^r a_j(\mu\times\nu)(E_j). \] The last equality is the lemma. Because the \(E_j\) are disjoint, the final sum equals \(\int_{X\times Y}f\,d(\mu\times\nu)\). For the reverse order, the same section representation gives \(\int_Xf(x,y)\,d\mu(x)=\sum_{j=1}^r a_j\mu(E_j^y)\); integrating over \(Y\) and applying the lemma to each \(E_j\) gives that same finite sum. This proves both identities. \(\square\)

Worked Computations

For the first iterated integral, if \(x\in A\), the section integral is \(2m(B)=2\cdot2=4\); if \(x\in C\), it is \(3m(D)=3\cdot1=3\); otherwise it is \(0\). Since \(m(A)=m(C)=1\), the outer integral is \(4\cdot1+3\cdot1=7\). In the other order, for \(y\in B\), the integral over \(X\) is \(2m(A)=2\); for \(y\in D\), it is \(3m(C)=3\); otherwise it is \(0\). The integral over \(Y\) is \(2m(B)+3m(D)=2\cdot2+3\cdot1=7\). Check: all three computations give \(7\), and both sections use the stated endpoint convention: \(A\) includes \(1\), whereas \(C\) does not.

For fixed \(y\in(0,1]\), \(E^y=[0,y)\), of measure \(y\); for \(y=0\), the section is empty and its measure is \(0\). Hence the other iterated integral is \(\int_0^1y\,dy=[y^2/2]_0^1=1/2\). The indicator identity gives \((m\times m)(E)=1/2\). Check: the two strict triangles \(x<y\) and \(y<x\) are exchanged by swapping coordinates and have equal area; the diagonal is null, since it is covered by \(n\) squares of side \(1/n\), of total area \(n(1/n^2)=1/n\). The unit square has area \(1\), so each triangle has area \(1/2\).

Integrating over \(Y\) first gives row sums \(1+2+0=3\) and \(3+0+4=7\); their sum over \(X\) is \(3+7=10\). Integrating over \(X\) first gives column sums \(1+3=4\), \(2+0=2\), and \(0+4=4\); their sum over \(Y\) is \(4+2+4=10\). Check: both iterated sums account for every matrix entry once, and each equals the product integral \(10\).

A Common Pitfall and the Scope of the Result

In a simple-function representation, the sets \(E_j\) must be disjoint if one applies the formula \(\int\sum_j a_j\chi_{E_j}=\sum_j a_j(\mu\times\nu)(E_j)\) in that form. For a concrete error, on \([0,1]^2\) let \(Q=[0,\tfrac12]\times[0,\tfrac12]\) and write \(f=\chi_Q+\chi_Q\). A mistaken treatment of the two identical sets as disjoint would give \((m\times m)(Q)+(m\times m)(Q)=\tfrac14+\tfrac14\) only if multiplicity is retained, but treating their union as one set with coefficient \(1\) gives the incorrect value \(\tfrac14\). The actual function is \(2\chi_Q\), so its product integral is \(2\cdot\tfrac14=\tfrac12\). For each \(x\in[0,\tfrac12]\), its section integral is \(2\cdot\tfrac12=1\), and it is \(0\) otherwise; the outer integral is \(1\cdot\tfrac12=\tfrac12\). The corrected approach is to combine coincident terms into one level set with the correct coefficient, or first rewrite the function using disjoint level sets.

The proof above depends on finite measures when it invokes uniqueness for finite product measures and ensures the indicator integrals are finite. It makes no claim here for arbitrary nonnegative measurable functions, whose values or integrals may be infinite. Extending the identity beyond simple functions requires an additional limiting argument; the simple-function result provides its finite-valued starting point. It also concerns the product sigma-algebra \(\mathcal{M}\otimes\mathcal{N}\), not an unmentioned completion.