Build a product measure from rectangles

Try

A rectangular region has width three and height two, so its area is six. If the height changes with horizontal position, area should accumulate the lengths of vertical sections. Before using that intuition, establish that sections are measurable and their lengths form a measurable function. Otherwise the proposed integral has not even been defined.

Predict

Predict whether every subset of a vertical line has a measurable vertical section. In the product sigma-algebra, yes. In the completed planar measure, not necessarily: all subsets of a null line become measurable in the plane, including a subset whose exceptional vertical section is nonmeasurable on the line. Completion strengthens the measurable-set collection and changes an every-point statement into an almost-everywhere one.

Experiment

For a region E, write E_x for the set of y whose pair (x, y) belongs to E. Compare these section profiles.

Region	Section length	Area
(0,3)×(0,2)	2 on (0,3)	6
0< y< x<1	x on (0,1)	1/2
0< x<1 and 0< y<1−x	1−x on (0,1)	1/2

Draw a vertical section at x=1/4 and at x=3/4. Explain which profile changes and which remains constant. These are fixed worked examples, not claims about every arbitrary subset of the plane.

Understand

Given measurable spaces (X, A) and (Y, B), the product sigma-algebra A⊗B is generated by measurable rectangles. Sections commute with complement and countable union. Thus the class of sets whose sections are measurable at every x is a sigma-algebra containing rectangles, proving section measurability for all product-measurable sets.

First assume μ and ν are finite measures. To show x↦ν(E_x) is measurable, consider the class D of product sets with that property. Rectangles belong because their section sizes are ν(B)1_A(x). The whole space belongs. Complements belong because their section sizes are ν(Y)−ν(E_x), with a finite first term. Disjoint countable unions belong because section sizes are limits of increasing sums of measurable functions. Therefore D is a lambda system containing the pi system of rectangles. The pi-lambda theorem implies that D contains the entire product sigma-algebra.

Define ρ(E)=integral ν(E_x)dμ(x). For disjoint E_j, the sections are disjoint and their measures add. Monotone convergence interchanges the increasing partial sums and the integral, proving countable additivity of ρ. On rectangles its value is μ(A)ν(B). This constructs the finite product measure directly from sections rather than assuming an unproved iterated-integral theorem.

For sigma-finite factors, partition each space into countably many disjoint measurable finite-measure pieces. Construct the finite product on each pair of pieces and sum the resulting measures. The rectangle formula follows by nonnegative summation. Uniqueness holds because any competing measure agrees on rectangle generators within each finite piece; the finite uniqueness theorem applies there, and summing the disjoint pieces proves global agreement. This argument also identifies precisely where sigma-finiteness enters.

Apply

Apply the section formula to a staircase made of finitely many rectangles, then to the triangular region in the table. Check that endpoints do not affect these length integrals. For a completed product measure, distinguish the original product-measurable representative from arbitrary modifications on null sets. Later theorems will allow such modifications while retaining conclusions only for almost every section.

Continue

Product measure is now a constructed measure with a rectangle rule and a uniqueness certificate. The next lesson extends the section formula from indicators to all nonnegative measurable functions. Keep track of the two spaces and measures separately; a counting-measure factor turns the same construction into a theorem about sums and integrals.

Continue to Interchange nonnegative iterated integrals.