# The Radon Nikodym derivative is non-zero almost everywhere

Let $\mu$ and $\nu$ be finite positive measures on a measure space and assume that $\nu$ is absolutely continuous with respect to $\mu$. That is, for every measurable set $E$, if $\mu(E) = 0$, then $\nu(E) = 0$. Via the Radon-Nikodym Theorem there must be a measurable function $h$ such that for all measurable sets $A$ we have $\nu(A) = \int_A h \text{ }d\mu$. This function $h$ is called the Radon-Nikodym derivative of $\nu$ with respect to $\mu$ and is denoted by $\frac{d\nu}{d\mu}$. Two questions:

1) With respect to the measure $\nu$, is $\frac{d\nu}{d\mu}$ almost everywhery non-zero?

2) With respect to the measure $\mu$, is $\frac{d\nu}{d\mu}$ almost everywhery non-zero?

• Take $\mu = m$ the Lebesgue measure and $\nu = 0$. Then ${d \nu \over d \mu} = 0$ ae. $[\mu]$. – copper.hat Aug 19 '17 at 15:55
• Yes, thank you, I agree! So this only leaves the first question. – Woett Aug 19 '17 at 16:00
• Compute $\nu(\{x: h(x)=0\})$. – John Dawkins Aug 19 '17 at 16:44