Question on integral, notation and Nikodym derivative

Question

I have a very general question for those measure theoric, real analysis guy out there . I am very confused by the concept of Nikodym derivative. If $v << \mu$, we can find a non negative function $f$ s.t: \begin{align} v(E) = \int_E fd\mu \end{align} We call $f$ the Radon-Nikodym derivative. So far I have no problem it's just notation. My book says that $f$ is sometimes written $\frac{dv}{d\mu}$ because it is the derivative of v w.r to $\mu$. Well this is again notation. What I don't get is why in the book they start to use this notation in calculus. Like simplifying fractions and even write : $fd\mu = dv$.

Isn't $\frac{dv}{d\mu}$ suppose to be just a notation ? If it's more than a notation what does it represent ? I understand the concept of function, measure, and integral but i'm lost when it comes to this notation. It has to represent something otherwise we won't be using it. And the fact that this fraction can be nicely used to derive the product rule for example doesn't comes for free.

So my question is clear, what does $\frac{dv}{d\mu}$ represent. The notion of derivative hasn't been introduced (yet) in the book so I am not supposed to conclude that if I derive v w.r to $\mu$ I get f.

Thanks for any help !!!

Answer 1

Remember from one-variable calculus that, given a (continuous) function $f:\mathbb{R}\to\mathbb{R}$, the indefinite integral of $f$ is the integral of $f$ (with respect to Lebesgue measure) without a domain of integration: $\int f(x)dx$. The Fundamental Theorem of Calculus says that any antiderivative $F$ of $f$ will give all the values $\int_a^bf(x)dx=F(b)-F(a)$, so we usually think that the indefinite integral is a family of functions (namely, $F(x)+k$ for $k\in\mathbb{R}$), but it is best to to think of indefinite integrals in the following way:

Given a measure space $(\Omega,\mathcal{A},\mu)$ and a measurable function $f:\Omega\to[0,\infty)$, we define the indefinite integral of $f$ (with respect to $\mu$) to be the measure $$(\int fd\mu):E\in\mathcal{A}\mapsto\int_E fd\mu$$ It is clear that $(\int fd\mu)<<\mu$. Also, for every $g$ integrable, the following formula holds: $$\int_X gd(\small{\int fd\mu}\normalsize)=\int_X gfd\mu$$

The Radon-Nikodym theorem gives the the converse of this construction: If $\nu<<\mu$, then there exists $f$ such that $\nu=(\int fd\mu)$. The Fundamental Theorem of Calculus gives us reason to think of $f$ as the derivative of $\nu$, so the notation $f=\dfrac{d\nu}{d\mu}$ is good for that purpose.

Now, let's use the fact above: if $g$ is integrable, then $$\int_X gd\nu=\int_X gd(\small \int fd\mu\normalsize)=\int_X gfd\mu$$ so this says that we can substitue $fd\mu$ by $d\nu$ inside integrals, or in a more concise manner:$d\nu=fd\mu$.

Answer 2

When we're integrating on the real line, and if $\nu$ and $\mu$ are identified with sufficiently smooth functions giving rise to Stieltjes measures, then in the case that $f$ is differentiable, it is literally true that $f = d\nu/d\mu$ in the sense of strong derivatives.

For example, suppose that $d\mu = dx$, the regular Lebesgue measure, and $d\nu$ is given the Stieltjes measure arising from $x^2$, so $d\nu([a,b]) = b^2 - a^2$. Then the Radon-Nikodym derivative is $f = d\nu/d\mu = d(x^2)/dx = 2x$.

Answer 3

Like you said, it is just a convenient notation. The measure $\nu$ is completely determined by the equation $$\nu(E)=\int_E\, d\nu=\int_E\, fd\mu, \quad \forall E\ \mu\text{-measurable}$$ which one usually writes $d\nu=fd\mu$ for short. Since $f$ depends only on $\mu$ and $\nu$, we write $$f=\frac{d\nu}{d\mu}, $$that's all.