Question

Suppose \((X, \mathcal{M}, \mu)\) is a \(\sigma\)-finite measure space and \(f \in L^{+}(X) .\) Let $$ G_{f}=\\{(x, y) \in X \times[0, \infty]: y \leq f(x)\\} $$ Then \(G_{f}\) is \(\mathrm{M} \times \mathrm{B}_{\mathrm{R}}\)-measurable and \(\mu \times m\left(G_{f}\right)=\int f d \mu ;\) the same is also true if the inequality \(y \leq f(x)\) in the definition of \(G_{f}\) is replaced by \(y

Short answer

\(G_f\) is measurable and \(\mu \times m(G_f) = \int f \, d\mu\), hence the area under \(f\) equals its integral.

Step-by-step solution

Understanding the Set G_f

The set \(G_{f}\) in this problem is defined as \(\{(x, y) \in X \times [0, \infty) : y \leq f(x)\}\). This represents a region under the graph of \(f(x)\) over the measure space \(X\). We are tasked with proving that \(G_{f}\) is measurable and that the measure of this set corresponds to the integral of \(f\) with respect to \(\mu\), effectively showing that the area under \(f\) corresponds to its integral.

Constructing the Map (x, y) -> f(x) - y

We consider the map \((x, y) \mapsto f(x) - y\). An important property of measurability would be that preimages of open sets are measurable. Observing that if \(y \leq f(x)\), then \(f(x) - y \geq 0\), we can relate \(G_f\) to the set where the map is non-negative.

Decomposing the Map into Composable Functions

The given map \((x, y) \mapsto f(x) - y\) can be decomposed into two functions: \((x, y) \mapsto (f(x), y)\) and \((z, y) \mapsto z - y\). The first function picks \(f(x)\) and \(y\) as coordinates, while the second evaluates the difference \(z - y\). This decomposition helps in demonstrating measurability by separately addressing the transformations.

Assessing Measurability of f(x) and (x,y)

The product space \(X \times [0, \infty)\) with the product \(\sigma\)-algebra \(\mathcal{M} \times \mathcal{B}_{\mathbb{R}}\) ensures that measurable functions like \(f(x)\) remain measurable when integrated with another measurable function, like \(y\). Measurability of \(f(x)\) implies \((x, y) \mapsto (f(x), y)\) is measurable, since \(y\) is constant for each \(x\).

Establishing the Integrability of f s.t. G_f Area Matches Integral

Finally, we assert that \(\mu \times m(G_f) = \int f \, d\mu\). Because \(\{(x, y): y \leq f(x)\}\) captures the vertical slices of height \(f(x)\), its measure in \(X \times [0, \infty)\) domain corresponds to the area under \(f\), consistent with the integral of \(f\) when \(f \in L^+(X)\). The product measure here verifies the theorem.

Key concepts

Sigma-finite Measure Space

A sigma-finite measure space refers to a measure space \((X, \mathcal{M}, \mu)\) where the entire space \(X\) can be broken down into a countable union of measurable subsets with a finite measure. In simpler terms, it's a flexible framework for dealing with infinite spaces by breaking them into manageable, finite pieces. This property is essential in measure theory as it allows us to extend results that are true in finite spaces to more complex, infinite spaces.

The concept of a sigma-finite measure space is vital in advanced calculus and analysis because it helps in defining integrals and measuring functions over complex spaces. In the context of our exercise, the sigma-finite nature of the measure space \((X, \mathcal{M}, \mu)\) allows us to think of the space as made up of smaller, finite parts. This makes it possible to utilize advanced integration techniques and theorems, such as Fubini's Theorem, which are applicable only in such structured spaces. This concept ensures that even though \(X\) can be infinitely large, calculations and integrations remain feasible.

Measurability

Measurability is a key concept in measure theory. It describes whether a set can be assigned a measure, which is essentially a generalized length or size. In the context of the exercise, the set \(G_f = \{(x, y) \in X \times [0, \infty) : y \leq f(x)\}\) must be shown to be measurable, meaning it can be assigned a measure in the product space \(X \times [0, \infty)\).

To demonstrate that \(G_f\) is measurable, we use the function \((x, y) \mapsto f(x) - y\). Here, measurability ensures that the preimages of open sets remain measurable. By breaking down the function into two parts, \((x, y) \mapsto (f(x), y)\) followed by \((z, y) \mapsto z - y\), we show it's compositions of measurable maps. This guarantees the original map is measurable, establishing \(G_f\) as measurable.

Understanding measurability is crucial because it helps confirm that operations, like integration or probability calculations, are valid. Without measurability, assigning a meaningful measure to \(G_f\) wouldn’t be possible, and thus the interpretation of the integral as an area wouldn't be feasible.

Integral of a Function

The integral of a function is a keystone concept in calculus. It represents the area under the curve of a function over an interval or a domain. In measure theory, especially in our context, the integral \(\int f \, d\mu\) seeks to find the total "size" or "mass" under the curve of \(f(x)\) weighted by the measure \(\mu\).

In the given exercise, showing that the measure \(\mu \times m(G_f) = \int f \, d\mu\) illuminates how integration equates to finding an area under the function \(f(x)\) in the measure space \((X, \mathcal{M}, \mu)\). This directly ties back to understanding how integrals, despite handling abstract and infinite inputs, can yield tangible geometric interpretations.

The concept is anchored in the notion from elementary calculus about finding the "area under the curve," but elevated into a more abstract framework. It allows integration over complex spaces and ensures the results are consistent and valid across various mathematical fields.