# Measure Theory and the Hahn Decomposition Theorem

Suppose that $\mu$ and $\nu$ are finite measures on a measurable space $(X,\mathcal{M})$. Prove that either $\nu \perp \mu$, or there exists $\epsilon > 0$ and a measurable set $E \subseteq X$ such that $\mu(E) > 0$ and $\nu \ge \epsilon \mu$ on $E$ (that is, every measurable set $S \subseteq E$ has $\nu(S) \ge \epsilon \mu(S)$).

Hint: Use Hahn Decomposition Theorem

Since $\mu$ and $\nu$ are measures, for $n\in \N$, $\nu-n^{-1}\mu$ is a signed measure. Then for each $n \in \N$ there exists a Hahn Decomposition for $X$ with respect to the signed measure $\nu-n^{-1}\mu$:

$X=P_{n}\cup N_{n}.$

Let $P=\cup_{n=1}^{\infty} P_{n}$ and $N=\cap_{n=1}^{\infty}N_{n}=P^{c}$. Then $N$ is negative for $\nu-n^{-1}\mu$ for every $n$. So we have

$(\nu-n^{-1}\mu)(N)\leq 0.$

Hence $\nu(N)-n^{-1}\mu(N)\leq 0$
and therefore, $0\leq \nu(N)\leq n^{-1} \mu(N).$
Since $\mu(N)<\infty$, letting $n\rightarrow \infty$ we have
$\nu(N)=0. (1)$
Now we consider two cases:

Case I: $\mu(P)=0$. Then from (1) we conclude $\mu\perp \nu$ .

Case II: $\mu(P)>0$. Then, $\exists \ \ n_{0}\in N$ such that $\mu(P_{n_{0}})>0$. Since $X=P_{n_0}\cup N_{n_0}$ is a Hahn decomposition with respect to $\nu-n_0^{-1}\mu$, $\nu(P_{n_{0}})-n_{0}^{-1} \mu(P_{n_{0}})>0,$ and hence $\nu(P_{n_{0}})>n_{0}^{-1} \mu(P_{n_{0}}).$
Letting $E:=P_{n_{0}}$ and $\epsilon=n_{0}^{-1}$ we have $\nu(E)>\epsilon \mu(E)$.       $\Box$