When is a function a Radon-Nikodym derivative

measure-theoryprobability theoryradon-nikodym

Or, stated another way, what requirements on $f$ makes $\mu(A) = \int_A f(x) \nu(x) dx$ a measure? (Here, $A$ is any measurable subset of $\mathbb{R}$ and $\nu$ is a measure on $\mathbb{R}$.)

I know the Radon-Nikodym theorem says that for any $\mu$ and $\nu$ that are measures on the same space, with $\mu \ll \nu$, there exists a Radon-Nikodym derivative $f$ for which the above holds, and furthermore $f$ is Borel-measurable. But for a given $\nu$, would the converse hold? I.e. for any Borel-measurable $f$, would the $\mu$ defined above be a measure? It does make sense that $\mu \ll \nu$, since if $\nu(A)=0$ for any $A$, then $\mu(A) = 0$ too.

Attempt at proof:

  1. for any $A \subseteq \mathbb{R}$, $\mu(A) \geq 0$ only if $f(x) \geq 0$ whenever $\nu(A) > 0$.
  2. $\mu(\varnothing) = 0$ since an integral over the empty set is always 0.
  3. for a disjoint set of sets $E_k \subset \mathbb{R}$, the measure of the union is $\mu(\cup_{k=1}^\infty E_k) = \sum_{k=1}^\infty \mu(E_k)$ by linearity of integrals.

So, it sounds like we need $f(x) \geq 0$ for all $x\in A$ such that \nu(A) > 0$. Is this the one requirement?

Best Answer

Some context:

If $(X, \mathcal{F}, \nu)$ is a measure space and $f : X \to [0, \infty]$ is measurable, then $$\mu(A) = \int_{A}f\, d\nu$$ defines a measure $\nu$ on $\mathcal{F}$, and $\mu \ll \nu$. This is straightforward to prove. This is commonly denoted $d\mu = f\,d\nu$ or $\mu(dx) = f(x)\nu(dx)$.

The Radon Nikodym theorem says that if $(X, \mathcal{F}, \nu)$ is $\sigma$-finite, then a converse is true: If $\mu \ll \nu$ is $\sigma$-finite, then there is a $\nu$-a.e. unique measurable $f : X \to [0, \infty)$ such that $d\mu = f\,d\nu$.

To answer your question, the necessary and sufficient condition for $d\mu = f\,d\nu$ to define a positive measure is that $f \geq 0$ a.e. $\nu$. The sufficiency was mentioned above. The necessity can be proved by contradiction: if $\nu(f < 0) > 0$, then $\nu(f < -h) > 0$ for some $h > 0$, so $\mu(f < -h) = \int_{\{f < -h\}}f\,d\nu \leq -h\nu(f < -h) < 0$, so $\mu$ is not a positive measure.