Interchange nonnegative iterated integrals

Try

A nonnegative surface can be approximated from below by stacks of simple measurable blocks. Add their volumes in either coordinate order. The result cannot suffer cancellation because every contribution is nonnegative. Tonelli’s theorem makes this geometric argument precise, even if the total volume is infinite. Finite volume is a conclusion to check, not a prerequisite of this theorem.

Predict

Predict the product integral of 1/(xy) on the open unit square. Both one-variable factors have infinite integrals, and Tonelli legitimately returns infinity. Now predict whether the same permission extends to a signed surface with positive and negative infinite contributions. It does not: rearranging cancellation can change iterated answers or leave the integral undefined.

Experiment

For the fixed surface f(x, y)=xy on the unit square, inspect horizontal and vertical accumulation.

First integration	Intermediate function	Final integral
Integrate y from 0 to 1	x/2	1/4
Integrate x from 0 to 1	y/2	1/4

Replace y by y² and recompute both columns. Then replace one factor by a nonnegative indicator to see how a geometric slice becomes a simple-function term. State the domain each time.

Understand

For sigma-finite measure spaces, Tonelli states that a nonnegative product-measurable function has measurable section-integral functions and that its product integral equals either iterated integral. Every value may be infinity. For an indicator, this is exactly the section construction of product measure, with the order-reversed construction agreeing by uniqueness. Finite nonnegative combinations establish the claim for simple functions.

For general f≥0, choose measurable simple s_n increasing to f. For each fixed x, monotone convergence in y gives integral s_n(x, y)dν(y) increasing to integral f(x, y)dν(y). The latter is measurable as an increasing limit of measurable functions. Applying monotone convergence in x shows that the limit of the iterated simple integrals equals the desired iterated integral. On the product space, monotone convergence gives the same limit as integral f. Repeating with the coordinates reversed completes the proof. The entire argument uses increasing nonnegative limits, so no undefined cancellation appears.

Counting measure is a useful factor: Tonelli becomes interchange of a nonnegative series and an integral. If the resulting integral of a sum of absolute values is finite, that sum is finite almost everywhere, furnishing a powerful convergence test.

Another consequence is the layer-cake identity. The set of pairs (x, t) with 0≤t< f(x) is product measurable: its strict inequality can be described using a countable union of rational intermediate thresholds. Its indicator integrates in t to f(x). Integrating in x first gives the measure of the upper level set where f>t. Tonelli therefore expresses integral f as the integral over t≥0 of μ(f>t). This translates a height profile into areas of horizontal layers.

Apply

For a nonnegative random variable, the layer-cake identity says its expectation is the integral of its survival probabilities. Derive the discrete analogue by counting unit layers for a nonnegative integer-valued variable. Be explicit about whether the threshold uses >k or ≥k, because shifting that convention shifts the summation index.

Continue

Tonelli is often the right first step for deciding whether a signed function is absolutely integrable: apply it to abs (f). If that integral is finite, Fubini will justify signed interchange. If it is infinite, do not infer that interchange is impossible in every special case; infer that this sufficient theorem no longer provides a certificate.

Continue to Check absolute integrability before changing order.