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

malalawak44 2022-07-12 Answered
Absolute continuity: For all ε > 0 there exists a δ > 0 such that if μ ( A ) < δ, A | f | d μ < 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 μ
Can I choose some special set A such that μ ( A ) < δ?
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Answers (1)

Kiley Hunter
Answered 2022-07-13 Author has 7 answers
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 > 0 we have M such that μ ( A M ) > μ ( E ) δ, so that μ ( A M c ) < δ. Now, if we choose δ so that μ ( B ) < δ 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 < .

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