malalawak44

2022-07-12

Absolute continuity: For all $\epsilon >0$ there exists a $\delta >0$ such that if $\mu (A)<\delta $, ${\int}_{A}|f|d\mu <\u03f5$. 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 (A)<\delta $?

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 (A)<\delta $?

Kiley Hunter

Beginner2022-07-13Added 7 answers

Let $f$ be finite almost everywhere, say, in $E$. Then if ${A}_{n}=\{x\mid |f(x)|\le M\}$, we have $\bigcup _{n=0}^{\mathrm{\infty}}{A}_{M}=E$.

Since it is an increasing sequence of sets ${A}_{M}\uparrow E$ and the measure is finite, for any $\delta >0$ we have $M$ such that $\mu ({A}_{M})>\mu (E)-\delta $, so that $\mu ({A}_{M}^{c})<\delta $. Now, if we choose $\delta $ so that $\mu (B)<\delta \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\int}_{B}|f|<1$, 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 < ∞ .

Since it is an increasing sequence of sets ${A}_{M}\uparrow E$ and the measure is finite, for any $\delta >0$ we have $M$ such that $\mu ({A}_{M})>\mu (E)-\delta $, so that $\mu ({A}_{M}^{c})<\delta $. Now, if we choose $\delta $ so that $\mu (B)<\delta \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\int}_{B}|f|<1$, we have

<math xmlns="http://www.w3.org/1998/Math/MathML" ">

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