Question

Let \((X, \mathcal{M}, \mu)\) and \((Y, \mathcal{N}, \nu)\) be arbitrary measure spaces (not necessarily \(\sigma\) finite). a. If \(f: X \rightarrow \mathbb{C}\) is \(\mathcal{M}\)-measurable, \(g: Y \rightarrow \mathbb{C}\) is N-measurable, and \(h(x, y)=\) \(f(x) g(y)\), then \(h\) is \(\mathcal{M} \otimes \mathcal{N}\)-measurable. b. If \(f \in L^{1}(\mu)\) and \(g \in L^{1}(\nu)\), then \(h \in L^{1}(\mu \times \nu)\) and \(\int h d(\mu \times \nu)=\) \(\left[\int f d \mu\right]\left[\int g d \nu\right]\).

Short answer

a. \(h\) is measurable. b. \( \int h d(\mu \times \nu) = \left(\int f d \mu\right) \left(\int g d \nu\right)\).

Step-by-step solution

Understand the Function Definition

We are given two functions: \(f: X \to \mathbb{C}\), which is \(\mathcal{M}\)-measurable, and \(g: Y \to \mathbb{C}\), which is \(\mathcal{N}\)-measurable. The function \(h: X \times Y \to \mathbb{C}\) is defined by \(h(x, y) = f(x)g(y)\). We need to show that this function is measurable with respect to the product \(\sigma\)-algebra \(\mathcal{M} \otimes \mathcal{N}\).

Use Properties of Measurable Functions

By definition, a function is \(\mathcal{M} \otimes \mathcal{N}\)-measurable if for every complex number \(c\), the set \(\{(x,y) \in X \times Y : h(x,y) = c\}\) is in \(\mathcal{M} \otimes \mathcal{N}\). Because \(f\) and \(g\) are measurable, their product \(h(x,y)\) is measurable with respect to the product \(\sigma\)-algebra (from the properties of measurable functions). Thus, \(h\) is \(\mathcal{M} \otimes \mathcal{N}\)-measurable.

Consider the Integrability Assumptions

We are given that \(f \in L^1(\mu)\) and \(g \in L^1(u)\), meaning they are integrable functions with respect to their respective measures. We must show that \(h(x,y) = f(x)g(y)\) is integrable with respect to the product measure \(\mu \times u\).

Apply Fubini's Theorem

According to Fubini's Theorem for product measures, if \(f \in L^1(\mu)\) and \(g \in L^1(u)\), then the product function \(h\) is integrable over the product measure space \((X \times Y, \mathcal{M} \otimes \mathcal{N}, \mu \times u)\). Furthermore, \[\int h \, d(\mu \times u) = \int \left(\int f(x)g(y) \, du(y) \right) d\mu(x) = \left(\int f \, d\mu\right)\left(\int g \, du\right).\] This gives us the desired result.

Key concepts

Measurable Functions

In measure theory, measurable functions play a crucial role in understanding how functions interact with measures. A function is considered measurable if it preserves the structure defined by measurable sets, essentially allowing for a meaningful way to integrate and analyze functions within a measure space. Specifically, a function \(f: X \to \mathbb{C}\) is \(\mathcal{M}\)-measurable if, for any real number \(a\), the set \(\{x \in X : f(x) > a\}\) belongs to \(\mathcal{M}\).
In our specific problem, we deal with two measurable functions: \(f: X \rightarrow \mathbb{C}\) being \(\mathcal{M}\)-measurable, and \(g: Y \rightarrow \mathbb{C}\) being \(\mathcal{N}\)-measurable. The product function \(h(x, y) = f(x)g(y)\) is of interest, and we determine its measurability with respect to the product \(\sigma\)-algebra \(\mathcal{M} \otimes \mathcal{N}\). Since \(f\) and \(g\) are measurable, their product, \(h\), is \(\mathcal{M} \otimes \mathcal{N}\)-measurable. It's a fundamental property that the pointwise product of two measurable functions results in another measurable function.

Fubini's Theorem

A cornerstone of measure theory is Fubini's Theorem, which provides a powerful tool for evaluating integrals over product measure spaces. When you integrate functions over such spaces, Fubini's Theorem allows you to simplify the process by breaking it into successive integrations over each individual space.
For functions \(f\) and \(g\) that are each in \(L^1\) over their respective measure spaces, Fubini’s Theorem asserts that their product \(h(x, y) = f(x)g(y)\) is integrable with respect to the product measure \(\mu \times u\). More importantly, it allows you to calculate
\[\int h\, d(\mu \times u) = \int \left(\int f(x)g(y) u(dy) \right) \mu(dx) = \left(\int f \, d\mu\right)\left(\int g \, du\right)\,
\]thereby making complex integrals more manageable by "slicing" them into simpler, iterated integrals.

• First, integrate \(g\) over \(Y\), with the measure \(u\).
• Next, integrate the result over \(X\), with the measure \(\mu\).

Product Measure Spaces

Understanding product measure spaces is crucial when dealing with two independent measure spaces simultaneously. In essence, when you consider two measure spaces, \((X, \mathcal{M}, \mu)\) and \((Y, \mathcal{N}, u)\), the product measure space \(X \times Y\), equipped with the \(\sigma\)-algebra \(\mathcal{M} \otimes \mathcal{N}\), allows you to examine coordinated phenomena across both dimensions.
Product measure spaces have structural characteristics derived from both individual spaces, effectively "combining" them into a comprehensive framework. This construction supports the formulation and integration of product function \(h(x, y) = f(x) g(y)\) as an \(\mathcal{M} \otimes \mathcal{N}\)-measurable function.
The product measure \(\mu \times u\) extends the concept of measure to the combined space, facilitating applications like Fubini's Theorem. In practice, this means you can handle more complex scenarios involving two independent variables as a unified entity, simplifying the analysis and computation of related integrals and other operations.