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Let $f:R\rightarrow R$ be a Borel measurable and assume $f$ is in $L_1$ [$0, T]$. Let $F(t)=\int$ $f(s) ds$. Show $d/dt$ $F(t)$ exists Lebesgue a.e on $[0, T]$ and is Borel measurable.

I am reading a book An introduction to Measure and Integration (second edition) by Rana, and on page 191 (Theorem 6.3.1) I can find a somewhat similar question, and the proof given there does make sense too. But the given proof is 2 pages long. He is using LDCT to supply the proof of the theorem.

I was wondering if there is a quick and short way of proving this result.

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Are you familiar with the Lebesgue differentiation theorem? This is almost an immediate consequence of the theorem. –  Ayman Hourieh Dec 20 '12 at 18:00
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