Absolute continuity: For all ε &gt; 0 there exists a δ &gt; 0 such that if μ ( A ) &lt; δ, ∫ A | f | d μ &lt; 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 μ is finite? ∫ | f | d μ = ∫ A | f | d μ + ∫ A c | f | d μCan I choose some special set A such that μ ( A ) &lt; δ?

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Absolute continuity: For all   ε  &amp;gt;  0 there exists a   δ  &amp;gt;  0 such that if   μ  (  A  )  &amp;lt;  δ,       ∫          A            |    f      |    d  μ  &amp;lt;  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   μ is finite?  ∫      |    f      |    d  μ  =      ∫    A        |    f      |    d  μ  +      ∫                  A        c                  |    f      |    d  μCan I choose some special set   A such that   μ  (  A  )  &amp;lt;  δ?

Kiley Hunter · Accepted Answer

Let   f be finite almost everywhere, say, in   E. Then if       A    n    =  {  x  ∣  |  f  (  x  )  |  ≤  M  }, we have       ⋃          n      =      0        ∞        A    M    =  E.Since it is an increasing sequence of sets       A    M    ↑  E and the measure is finite, for any   δ  &amp;gt;  0 we have   M such that   μ  (      A    M    )  &amp;gt;  μ  (  E  )  −  δ, so that   μ  (      A    M    c    )  &amp;lt;  δ. Now, if we choose   δ so that   μ  (  B  )  &amp;lt;  δ    ⟹        ∫    B    |  f  |  &amp;lt;  1, we have&amp;lt;math xmlns=&quot;http://www.w3.org/1998/Math/MathML&quot; &quot;&amp;gt;  ∫  |  f  |  =      ∫    E    |  f  |  =      ∫                  A        M              |  f  |  +      ∫                  A        M        c              |  f  |  ≤  M  μ  (      A    M    )  +  1  &amp;lt;  ∞  .

Absolute continuity: For all ε > 0 there exists a δ<!-- δ

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