I am proving something about uniform integrable families of functions and I need to prove something that seems easy but I stucked, it says:
Let be $(X,S,\mu)$ a measurable space, $g$ integrable and $\epsilon>0$ so exists $\delta>0$ s.t. If $E\in S$ with $\mu(E)<\delta$ then $\int_{E}|g|d\mu<\epsilon.$
You're trying to prove that the measure induced by $|g|$ is absolutely continuous.
Given $\epsilon >0$, there is some simple function $\phi = \sum_{i=1}^n a_i \chi_{A_i}$ such that $0\leq \phi\leq g$ and $\int \phi \geq \int g - \frac{\epsilon}2$
Note that $$\begin{align} \int_{E} g &= \int_E (g-\phi) + \int_E\phi\\ &\leq \int (g-\phi) + \sum_{i=1}^n a_i \mu(E\cap A_i)\\ &\leq \frac{\epsilon}2 + n\cdot \max a_i \cdot \mu(E) \end{align}$$
Therefore $\delta = \frac{\epsilon}{2n\max a_i}$ fits the bill.