The following question was asked in an assignment of Measure theory and I am struggling to solve it.
Question :Let $\mu$ be a lebesgue measure on $\mathbb{R}$ and m be the counting measure.(1) Then show that $\mu$ is absolutely continuous wrt to m.(2) Can we find a function $h$ such that $\mu =hm?$.
We say that $\lambda$ is absolutely continuous wrt $\mu$ if $\lambda (E)=0$ for every $E\in M$ for which $\mu(E)=0$.
If the counting measure of X is 0. Then the number of elements in the set is 0. Then the lebesgue measure is trivially 0.
But I am not able to make any progress regarding 2nd part. Can you please help me with the question?
Best Answer
I assume your notation is meant to indicate $\mu(A) = \int_A h \, dm$.
Assume for sake of contradiction that such a function $h$ exists. For any singleton set $\{x\}$, we have $\mu(\{x\})=0$ and $m(\{x\})=1$ which implies $h(x)=0$. Since $x$ was arbitrary, we have $h \equiv 0$, which contradicts the fact that the Lebesgue measure is nonzero.
Note that this shows why the $\sigma$-finite condition is needed for the Radon-Nikodym theorem, in the sense that $\mu$ being absolutely continuous with respect to $m$ (what you showed in part 1) is not enough to ensure the existence of a Radon-Nikodym derivative $h$.