Let $\nu$ be a signed measure which is absolutely continuous to a sigma-finite measure $\mu$. Show that $\frac{d|\nu|}{d\mu}=|\frac{d\nu}{d\mu}|$, where $|\nu|$ is the total variation measure of $\nu$.

Now by definition, for any $E\in F$ we have $|\nu|(E)=\nu^+(E)+\nu^-(E)=\nu(E\cap P)-\nu(E\cap P^c)$, where $\{P,P^c\}$ is a Hahn decomposition of $\nu$ and $\{\nu^+,\nu^-\}$ is the corresponding Jordan decomposition. Using these two expressions, I found that $\frac{d|\nu|}{d\mu}=\frac{d\nu^+}{d\mu}+\frac{d\nu^-}{d\mu}=(1_P-1_{P^c})\frac{d\nu}{d\mu}$, but I'm not sure how to equate either of these expressions with $|\frac{d\nu}{d\mu}|$.

  • Are you looking for proof of $\frac{d|\nu|}{d\mu}=|\frac{d\nu}{d\mu}|$. – UserS Dec 6 at 16:03
  • @UserS Yes, I am. – Keshav Srinivasan Dec 6 at 16:11
up vote 1 down vote accepted

Note that $|\nu|$ is a measure i.e. non-negative valued. Now if you go through the proof of Radon-Nikodym theorem you see that proof has two parts, one is for (non-negative) measure in which case Radon-Nikodym derivative is non-negative and another part is for signed measure which has been done by considering positive and negative parts of signed measure giving extended real valued Radon-Nikodym derivative. Therefore $\frac{d|\nu|}{d\mu}\geq 0$ and so taking modules in the equation $\frac{d|\nu|}{d\mu}=\frac{d\nu^+}{d\mu}+\frac{d\nu^-}{d\mu}=(1_P-1_{P^c})\frac{d\nu}{d\mu}$ we have $\frac{d|\nu|}{d\mu}=|(1_P-1_{P^c})||\frac{d\nu}{d\mu}|=1|\frac{d\nu}{d\mu}|=| \frac{d\nu}{d\mu}|$ , $\mu$ a.e.

Let $P, N$ be the Hahn decomposition of $\nu$, with the associated Jordan decomposition $\nu^+, \nu^-$. One sees that $\nu << \mu \implies \nu^+ << \mu$ and $\nu^- << \mu$. Let $f = \frac{d \nu}{d \mu}$.

These are both positive measures, and therefore admit Radon-Nikodym derivatives with respect to $\mu$, which we denote by $g = \frac{d \nu^+}{d \mu}$ and $h = \frac{d\nu^-}{d \mu}$. Note that $g,h$ are non-negative.

From the above, it follows that $\nu^- + \nu^+ = |\nu| << \mu$.

First, compute $g,h$ in terms of $f$. For any function $K$, we have : $$\int Kf d \mu = \int K d \nu = \int_P K d \nu^+ - \int_{P^C} K d \nu^- = \int K \left(g1_P - h1_{P^C}\right)d \mu$$

So, $f = g1_P - h1_{P^c}$ almost everywhere, by uniqueness of the RN derivative.

Task : Use a similar argument to above, to show that $\frac{d|\nu|}{d \mu} = g1_P + h1_{P^c}$. Conclude that $|f| = \frac{d |\nu|}{d \mu}$(since $P^c$ and $P$ are disjoint, the absolute value of $g1_P - h1_{P^c}$ is simply equal to $g1_P + h1_{P^c}$, as desired).

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.