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.

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.
