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Finite products of measures with densities have the product density

Statement

Let $μ$ be any finite measure on $(W, B_{W})$ , where $W$ and its trace Borel sigma-algebra $B_{W}$ are as in Finite-window spatial observations and configurations. Let $f : W \to [0, \infty]$ be measurable with $\int_{W} f d μ < \infty$ , and put $ν = f d μ$ . For every integer $n \geq 1$ , the product measure $ν^{\otimes n}$ on $B_{W}^{\otimes n}$ has density $F_{n} (x_{1}, \dots, x_{n}) = \prod_{i = 1}^{n} f (x_{i}),$ where products use the extended-product convention $0 \cdot (+ \infty) = 0$ . For $n = 0$ , both product measures are the unit mass on the one-point empty tuple and the product density is $1$ .

Facts & Assumptions

Given: A finite measure $μ$ , a nonnegative measurable $f$ with finite integral, and its density measure $ν = f d μ$ .

[F1]

The set function $A \mapsto \int_{A} f d μ$ is a measure. (The measure with density $f$ relative to $μ$ )

[F2]

Finite products of sigma-finite measures exist and have the rectangle product formula. (For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique)

[F3]

Tonelli's theorem interchanges nonnegative iterated integrals on a sigma-finite product. (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product)

[F4]

Two finite measures agreeing on a generating pi-system that contains the whole space agree on the generated sigma-algebra: the constant sequence equal to the whole space is a common finite-measure exhaustion. (Measures agreeing on a generating pi-system are equal under an increasing finite-measure exhaustion from that pi-system)

[F5]

Finite products of nonnegative measurable functions are measurable. For two such functions $u, v$ , for $a \geq 0$ , ${u v > a} = ⋃_{q \in Q, q > 0} {u > q} \cap {v > a / q},$ and for $a < 0$ the superlevel set is the whole space. Iterating this identity proves the finite-product claim, using $0 \cdot (+ \infty) = 0$ . (Extended-real-valued measurable functions)

Proof

Given: $n \geq 1$ , $μ (W) < \infty$ , and $\int_{W} f d μ < \infty$ .

1.1givenF1F3F5construct

Each coordinate projection $π_{i} : W^{n} \to W$ is measurable because inverse images of members of $B_{W}$ are measurable rectangles. Hence each $f \circ π_{i}$ is measurable, and [F5] makes $F_{n} = \prod_{i = 1}^{n} f \circ π_{i}$ measurable. Thus $A \mapsto \int_{A} F_{n} d μ^{\otimes n}$ is a measure by the same indefinite-integral construction as [F1]. Its total mass is $(\int_{W} f d μ)^{n} < \infty$ by repeated Tonelli [F3].

1.2givenF1F2F3algebra

For a measurable rectangle $B_{1} \times \dots \times B_{n}$ , repeated use of [F2] gives $ν^{\otimes n} (B_{1} \times \dots \times B_{n}) = \prod_{i} ν (B_{i}) = \prod_{i} \int_{B_{i}} f d μ$ . Repeated Tonelli [F3] gives the same value for $\int_{B_{1} \times \dots \times B_{n}} F_{n} d μ^{\otimes n}$ .

2.1step 1.1step 1.2F2F4∎

Rectangles form a generating pi-system containing the whole space $W^{n}$ , and the product formula for $W^{n}$ in step 1.2 shows that the two finite measures have the same finite total mass. Therefore [F4] applies and gives equality on all of $B_{W}^{\otimes n}$ . For $n = 0$ the statement is the definition of the empty product measure.

Depends on

Used by

Finite-window spatial Poisson likelihood relative to unit rate Theorem

Dependency tree · two levels

42 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Adrian Baddeley, Spatial Point Processes and their Applications, complete lecture notes (standard reference, not scraped)
Günter Last and Mathew Penrose, Lectures on the Poisson Process, author-hosted 2017 version (standard reference, not scraped)