Radiation of Measures and the Radon-Nikodym Theorem

Introduction

The Radon-Nikodym Theorem is a fundamental theorem in measure theory that characterizes the absolute continuity of measures. The theorem states that if two measures $\\mu$ and $\u$ are defined on the same measurable space, and $\u$ is absolutely continuous with respect to $\\mu$, then there exists a unique measurable function $f$ such that $\u(E) = \\int_E f d\\mu$ for all measurable sets $E$. This function $f$ is known as the Radon-Nikodym derivative of $\u$ with respect to $\\mu$. In this article, we will explore a special case of the Radon-Nikodym Theorem, namely the radiation of measures.

Radiation of Measures

Let $(\\Omega, \\mathcal{F})$ be a measurable space and $\\mu$ be a measure defined on $\\mathcal{F}$ such that $\\mu(\\Omega) = \\infty$. We call a measure $\u$ on $(\\Omega, \\mathcal{F})$ a radiation of $\\mu$ if $\u(E) = \\mu(E \\cap A)$ for some $A \\subset \\Omega$ with $\\mu(A) = \\infty$ and for all $E \\in \\mathcal{F}$. Intuitively, we can think of $\u$ as the measure of $E$ when we only consider the part of $\\Omega$ that intersects with $A$. We can show that any radiation of $\\mu$ is absolutely continuous with respect to $\\mu$. To see this, suppose $\u$ is a radiation of $\\mu$ and $E \\in \\mathcal{F}$ is such that $\\mu(E) = 0$. Then $\\mu(E \\cap A) = 0$ since $E \\cap A \\subset E$ and $\\mu(A) = \\infty$. Hence $\u(E) = 0$ and we have $\u \\ll \\mu$.

The Radon-Nikodym Theorem for Radiation of Measures

Now we can state and prove the special case of the Radon-Nikodym Theorem for radiation of measures.Theorem: Let $\\mu$ be a measure on $(\\Omega, \\mathcal{F})$ such that $\\mu(\\Omega) = \\infty$, and let $\u$ be a radiation of $\\mu$. Then there exists a unique measurable function $f$ such that $\u(E) = \\int_E f d\\mu$ for all measurable sets $E$.Proof: Let $A \\subset \\Omega$ be such that $\\mu(A) = \\infty$ and $\u(E) = \\mu(E \\cap A)$ for all $E \\in \\mathcal{F}$. Since $\u \\ll \\mu$, we have $\u(E) = 0$ whenever $\\mu(E) = 0$. Let $f = \\frac{d\u}{d\\mu}$. Then for any $E \\in \\mathcal{F}$, we have\\begin{align*}\\int_E f d\\mu &= \\int_{E \\cap A} f d\\mu \\\\&= \\int_{E \\cap A} \\frac{d\u}{d\\mu} d\\mu \\\\&= \\int_{E \\cap A} 1 d\u \\\\&= \u(E \\cap A) \\\\&= \u(E).\\end{align*}Thus $f$ satisfies the desired property. To show the uniqueness of $f$, suppose $g$ also satisfies $\u(E) = \\int_E g d\\mu$ for all measurable sets $E$. Then for any $E \\in \\mathcal{F}$, we have\\begin{align*}\\int_E (f-g) d\\mu &= \\int_E f d\\mu - \\int_E g d\\mu \\\\&= \u(E) - \u(E) \\\\&= 0.\\end{align*}Since $\\mu(\\Omega) = \\infty$, we can choose a sequence $\\{E_n\\}$ of disjoint measurable sets in $\\mathcal{F}$ such that $\\mu(E_n) \\rightarrow \\infty$ as $n \\rightarrow \\infty$. Then\\begin{align*}\\int_{\\Omega} |f-g| d\\mu &= \\sum_{n=1}^\\infty \\int_{E_n} |f-g| d\\mu \\\\&\\geq \\sum_{n=1}^\\infty (\\inf_{E_n} (f-g)) \\mu(E_n) \\\\&= \\infty.\\end{align*}Hence $f = g$ almost everywhere with respect to $\\mu$, and the proof is complete.

Conclusion

In this article, we have explored the special case of the Radon-Nikodym Theorem for radiation of measures. We have shown that any radiation of a measure $\\mu$ is absolutely continuous with respect to $\\mu$, and that there exists a unique measurable function $f$ such that $\u(E) = \\int_E f d\\mu$ for all measurable sets $E$. This result has important applications in probability theory, where radiation of measures is used to model the behavior of random processes that can have infinite activity.