If $\nu \ll \mu$, there exists $E \in \cal F$ and $n \geq 1$ such that $\nu(E) >0$ and $n^{-1} \mu(A) \leq \nu(A) \leq n\mu(A)$ for all $A \subset E$. I'm thinking about the following question from an old prelim:
Let $(X, \cal{F})$ be a measure space with $\sigma$-finite measures $\mu$ and $\nu$ (it does not say whether these are signed measures). If $\nu \ll \mu$ and $\nu \neq 0$, then there exists $E \in \cal F$ and $n\in \mathbb{N}$ such that $\nu(E) >0$ and $n^{-1} \mu(A) \leq \nu(A) \leq n\mu(A)$ for all $A \in \cal{F}$ with $A \subset E$. 
Let's assume that these are positive measures. Since both are $\sigma$-finite and $\nu \ll \mu$, we can choose and extended $\mu$-integrable function $f:X \to \mathbb{R}$ such that $d\nu = fd\mu$. Then we hope to find $n\geq 1$ and $E \in \cal{F}$ such that 
$$
n^{-1} \mu(A) \leq \int_A f d\mu \leq n\mu(A)
$$
for all measurable $A \subset E$. Since $n \mu(A) = \int_A n d\mu$ and $n^{-1}\mu(A) = \int_A n^{-1} d\mu$, I thought about looking at the set $E_n= \{ x \in X: n^{-1} \leq f(x) \leq n\}$ for appropriately chosen $n$, but I'm not sure if this is the right idea. 
 A: Yes, your approach is fine. Pick some set $X_0 \in X$ such that $0<\nu(X_0)<\infty$ (such a set exists because $\nu$ is $\sigma$-finite and $\nu \neq 0$), and consider
$$E_n := \{x \in X; n^{-1} \leq f(x) \leq n\} \cap X_0.$$
Clearly, $\nu(E_n)<\infty$ for all $n \in \mathbb{N}$ and $E_n \uparrow F := \{x \in X; 0<f(x)<\infty\} \cap X_0$, and so it follows from the continuity of the measure $\nu$ that $\lim_{n \to \infty} \nu(E_n)=\nu(F)$. If we can show that $\nu(F)>0$, then it follows that $\nu(E_n)>0$ for large $n$, and we are done.
Suppose, to the contrary, that $\nu(F)=0$, then
\begin{align*} \nu(X_0) &= \nu(X_0 \cap \{x \in X; f(x) =0 \, \text{or} \, f(x)=\infty\}) \\ &= \underbrace{\int_{X_0} f(x) 1_{\{f(x)=0\}} \, \mu(dx)}_{0} + \int_{X_0} f(x) 1_{\{f(x)=\infty\}} \, \mu(dx)\end{align*}
This, however, contradicts our choice of $X_0$, i.e. the fact that the left-hand side is a finite number in $(0,\infty)$. (The right-hand side cannot be in $(0,\infty)$; it is either $0$ or $\infty$.)
