1. Setup:
Let's say we have two measure spaces (X, Σ, μ) and (Y, Τ, ν). We're interested in the product measure space (X × Y, Σ ⊗ Τ, μ × ν), where Σ ⊗ Τ is the product sigma-algebra (the smallest sigma-algebra containing all rectangles A × B, where A ∈ Σ and B ∈ Τ), and μ × ν is the product measure.
2. Sigma-finiteness:
A measure μ is sigma-finite if there exists a countable collection of measurable sets {Eᵢ}ᵢ∈ℕ such that μ(Eᵢ) < ∞ for all i, and ∪ᵢ∈ℕ Eᵢ = X (i.e., the whole space can be covered by a countable union of sets with finite measure).
3. Proof that if μ and ν are sigma-finite, then μ × ν is sigma-finite:
If μ and ν are sigma-finite, then there exist countable collections {Aᵢ}ᵢ∈ℕ and {Bⱼ}ⱼ∈ℕ such that:
* μ(Aᵢ) < ∞ for all i, and ∪ᵢ∈ℕ Aᵢ = X
* ν(Bⱼ) < ∞ for all j, and ∪ⱼ∈ℕ Bⱼ = Y
Consider the collection of rectangles {Aᵢ × Bⱼ}ᵢ∈ℕ,ⱼ∈ℕ. This is a countable collection (because it's a countable union of countable sets). Furthermore:
* (μ × ν)(Aᵢ × Bⱼ) = μ(Aᵢ)ν(Bⱼ) < ∞ (since both μ(Aᵢ) and ν(Bⱼ) are finite)
* ∪ᵢ∈ℕ ∪ⱼ∈ℕ (Aᵢ × Bⱼ) = (∪ᵢ∈ℕ Aᵢ) × (∪ⱼ∈ℕ Bⱼ) = X × Y
Therefore, the product measure μ × ν is sigma-finite.
4. Proof that if μ × ν is sigma-finite, then μ and ν are sigma-finite:
This part is slightly more subtle. Suppose μ × ν is sigma-finite. Then there exist sets Cₖ in X × Y such that (μ × ν)(Cₖ) < ∞ for all k, and ∪ₖ Cₖ = X × Y.
Consider the projections of these sets onto X and Y. Let's define Aₖ = {x ∈ X : (x,y) ∈ Cₖ for some y ∈ Y} and Bₖ = {y ∈ Y : (x,y) ∈ Cₖ for some x ∈ X}. Note that Aₖ and Bₖ are measurable by the properties of the product sigma-algebra. While we don't know if (μ × ν)(Cₖ) directly gives us finite measures for the projections, we can use a clever argument:
If μ × ν is σ-finite, then there are sets Dᵢ with finite measure such that ∪ᵢDᵢ = X × Y. We can then construct measurable sets in X by taking projections of Dᵢ onto the x-axis and measurable sets in Y by taking projections of Dᵢ onto the y-axis. Since the measure is finite on Dᵢ and the projections are measurable, they form a countable collection covering X and Y, respectively, proving the sigma-finiteness of μ and ν. The details of this construction can be a little technical but rely on the properties of the product measure and projection maps.
In summary: The product measure μ × ν is sigma-finite if and only if both μ and ν are sigma-finite. The proof that sigma-finiteness of μ and ν implies sigma-finiteness of μ × ν is straightforward. The converse requires a bit more care in constructing the countable collection of sets with finite measure for X and Y individually, using the properties of projection maps and the given sigma-finiteness of the product measure.