Radon-Nikodym Derivative of a Total Variation Measure

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 '18 at 16:03
• @UserS Yes, I am. – Keshav Srinivasan Dec 6 '18 at 16:11

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).