Application of Radon Nikodym Theorem on Absolutely Continuous Measures I have the following problem:
Show $\beta \ll \eta$  if and only if for every $\epsilon > 0 $ there exists a $\delta>0$ such that $\eta(E)<\delta$ implies $\beta(E)<\epsilon$.
For the forward direction I had a proof, but it relied on the use of the false statement that "$h$ integrable implies that $h$ is bounded except on a set of measure zero".
I had no problem with the backward direction.
 A: No need to use Radon-Nikodým here. I'll assume that $\beta$ is totally finite. See the end of the answer why this is necessary.
Suppose that $\beta E = 0$ whenever $E$ is measurable and $\eta E = 0$ but that the desired $\varepsilon$-$\delta$-condition doesn't hold.
Then there is $\varepsilon \gt 0$ such that for all $\delta \gt 0$ there is $E$ such that $\eta(E) \lt \delta$ but $\beta(E) \geq \varepsilon$. For each $n$ choose $E_n$ such that $\eta(E_n) \lt 2^{-n}$ and $\beta(E_n) \geq \varepsilon$. 
Define $E = \bigcap_{N \in \mathbb{N}} \bigcup_{n \geq N} E_n$. Then
$$
0 \leq \eta E \leq \inf_{N \in \mathbb N} \eta \bigcup_{n \geq N} E_n \leq \inf_{N \in \mathbb{N}} \sum_{n \geq N} 2^{-n} = 0
$$
hence $\eta E  = 0$. By hypothesis it follows that $\beta E = 0$ as well. On the other hand, assuming $\beta$ is totally finite, we get
$$
0 = \beta E = \lim_{N \to \infty} \beta \bigcup_{n \geq N} E_n \geq \varepsilon \gt 0,
$$
which is absurd.

Note that, as @copper.hat pointed out in the comments, it is necessary to assume that $\beta$ is totally finite.  The $\sigma$-finite example $\beta E = \int_{E} \frac{1}{t}\,dt$ on $(0,1)$ shows this: For Lebesgue measure $\lambda$ on $(0,1)$, the absolute continuity condition “$\lambda E = 0$ implies that $\beta E = 0$” holds, while the $\varepsilon$-$\delta$-characterization doesn't. For every $\delta \gt 0$ we have $\beta(0,\delta) = \infty$ while $\lambda(0,\delta) = \delta$.
A: Assume that $\beta=h\eta$ with $h\geqslant0$ integrable with respect to $\eta$, in particular $\beta$ is a finite measure. Let $\varepsilon\gt0$. 
There exists some finite $t_\varepsilon$ such that $\beta(B_\varepsilon)=\int_{B_\varepsilon} h\,\mathrm d\eta\leqslant\varepsilon$ where $B_\varepsilon=[h\geqslant t_\varepsilon]$. Note that, for every measurable $A$, $A\subset B_\varepsilon\cup(A\setminus B_\varepsilon)$, hence $\beta(A)\leqslant\beta(B_\varepsilon)+\beta(A\cap[h\leqslant t_\varepsilon])\leqslant\varepsilon+t_\varepsilon\eta(A)$.
Let $\delta=\varepsilon/t_\varepsilon$. One sees that, for every measurable $A$, if $\eta(A)\leqslant\delta$, then $\beta(A)\leqslant2\varepsilon$, QED.
