91Ó°ÊÓ

Es seien \(\mu\) ein \(\sigma\)-endliches \(\mathrm{MaB}\) und \(\nu, \rho\) signierte Maße auf \(\mathfrak{A}\). a) Aus \(\nu \ll \mu\) folgt \(\alpha \nu \ll \mu(\alpha \in \mathbb{R})\) und \(d(\alpha \nu) / d \mu=\alpha(d \nu / d \mu) \mu\)-f.ü. b) Nehmen \(\nu\) und \(\rho\) beide den Wert \(+\infty\) oder beide den Wert \(-\infty\) nicht an, so ist \(\nu+\rho \ll \mu\), falls \(\nu \ll \mu\) und \(\rho \ll \mu\), und dann gilt: \(d(\nu+\rho) / d \mu=d \nu / d \mu+d \rho / d \mu \mu\)-f.ü. c) Ist \(\nu\) ein \(\sigma\)-endliches Maß mit \(\rho \ll \nu, \nu \ll \mu\), so gilt die „Kettenregel" $$ \frac{d \rho}{d \mu}=\frac{d \rho}{d \nu} \frac{d \nu}{d \mu} \quad \mu \text {-f.ü. } $$ d) Ist \(\nu\) ein \(\sigma\)-endliches Maß mit \(\nu \ll \mu\) und \(\mu \ll \nu\), so ist \(d \nu / d \mu \neq 0 \mu\)-f.ü. und $$ \frac{d \nu}{d \mu}=\left(\frac{d \mu}{d \nu}\right)^{-1} \quad \mu \text {-f.ü. } $$

Step by step solution

01

Understanding the Proofs

This exercise contains multiple statements about Radon-Nikodym derivatives, each of which represents a different property of measures and needs to be proven. Each part of the exercise must be analyzed individually, thereby leading to different proofs. Let's start with part a) and then build up to parts b), c), and d).

02

Proving Part a)

We are given that \(\nu \ll \mu\), which means there is a nonnegative measurable function \(f\) (\(f = \frac{d\nu}{d\mu}\)) such that \(\nu(A) = \int_A f d\mu\) for all measurable sets \(A\). If we multiply \(\nu\) by a real number \(\alpha\), we obtain a new measure \(\alpha \nu(A) = \alpha \int_A f d\mu = \int_A \alpha f d\mu\). Thus \(\alpha \nu \ll \mu\), and \(\frac{d(\alpha \nu)}{d\mu} = \alpha \frac{d\nu}{d\mu} = \alpha f\), \(\mu\)-almost everywhere.

03

Proving Part b)

Assume \(\nu \ll \mu\) and \(\rho \ll \mu\), and neither \(\nu\) nor \(\rho\) take on the values \(\pm\infty\). Then, we have that \(\nu + \rho (A) = \int_A f d\mu + \int_A g d\mu = \int_A (f + g) d\mu\), with \(f = \frac{d\nu}{d\mu}\) and \(g = \frac{d\rho}{d\mu}\). Hence \(\nu + \rho \ll \mu\) and \(\frac{d(\nu + \rho)}{d\mu} = \frac{d\nu}{d\mu} + \frac{d\rho}{d\mu} = f + g\), \(\mu\)-almost everywhere.

04

Proving Part c)

If \(\rho \ll \nu\) and \(\nu \ll \mu\), we can use the chain rule for Radon-Nikodym derivatives to write \(\frac{d\rho}{d\mu} = \frac{d\rho}{d\nu} \frac{d\nu}{d\mu}\), \(\mu\)-almost everywhere. This equation is valid under the assumption that \(\nu\) is a \(\sigma\)-finite measure.

05

Proving Part d)

If \(\nu \ll \mu\) and \(\mu \ll \nu\), it's known that the Radon-Nikodym derivative \(\frac{d\nu}{d\mu}\) exists and is non-zero \(\mu\)-almost everywhere, because \(\nu\) is absolutely continuous with respect to \(\mu\). Applying the inverse function, we get that \(\frac{d\nu}{d\mu} = \left(\frac{d\mu}{d\nu}\right)^{-1}\), \(\mu\)-almost everywhere.

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