# Absolute continuity: For all &#x03B5;<!-- ε --> &gt; 0 there exists a &#x03B4;<!-- δ

Absolute continuity: For all $\epsilon >0$ there exists a $\delta >0$ such that if $\mu \left(A\right)<\delta$, ${\int }_{A}|f|d\mu <ϵ$. Here $A$ is a measurable subset of $E$.
I know that if $f$ is integrable then it is absolutely continuous. But is there anyway I can show that absolute continuity implies integrability when $\mu$ is finite?
$\int |f|d\mu ={\int }_{A}|f|d\mu +{\int }_{{A}^{c}}|f|d\mu$
Can I choose some special set $A$ such that $\mu \left(A\right)<\delta$?
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Kiley Hunter
Let $f$ be finite almost everywhere, say, in $E$. Then if ${A}_{n}=\left\{x\mid |f\left(x\right)|\le M\right\}$, we have $\bigcup _{n=0}^{\mathrm{\infty }}{A}_{M}=E$.
Since it is an increasing sequence of sets ${A}_{M}↑E$ and the measure is finite, for any $\delta >0$ we have $M$ such that $\mu \left({A}_{M}\right)>\mu \left(E\right)-\delta$, so that $\mu \left({A}_{M}^{c}\right)<\delta$. Now, if we choose $\delta$ so that $\mu \left(B\right)<\delta \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{\int }_{B}|f|<1$, we have
<math xmlns="http://www.w3.org/1998/Math/MathML" "> | f | = E | f | = A M | f | + A M c | f | M μ ( A M ) + 1 < .