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 Dec15 comment Absolutely continuous functions with derivatives in $L^p$ @Jonas. Yes you are totally right. It is integrable on bounded intervals, and this is what I use for the proof. Sorry for my confusion! Dec15 comment Absolutely continuous functions with derivatives in $L^p$ @Jonas. Yes thank you! Dec15 comment Absolutely continuous functions with derivatives in $L^p$ ...is integrable, and we have $F(b)-F(a)= \int_a^b F'(x)dx$. I will post an answer to my question so that you will understand the discussion between Giuseppe and I. Dec15 comment Absolutely continuous functions with derivatives in $L^p$ @Jonas: The problem comes from a problems set that our professor gave us to practise for the final exam. I am not sure where she took them from. The definition of absolutely continuous function that I use is: For every $\epsilon>0$, there exists $\delta > 0$ such that for any finite collection of disjoint intervals $(a_i, b_i)_{i=1}^n$, $\sum_i |F(b_i)-F(a_i)|<\epsilon$ whenever $\sum_i (b_i-a_i) < \delta$. This is the definition from Folland's real analysis. The Fundamental theorem of calculus states that $F$ is absolutely continuous if and only if $F'$ exists almost everywhere and... Dec14 comment Absolutely continuous functions with derivatives in $L^p$ @Jonas. Is it important in this case? Dec14 comment Absolutely continuous functions with derivatives in $L^p$ @Jonas. It is not directly part of my definition, but it follows from the fundamental theorem of calculus for Lebesgue integration. Dec12 comment Absolutely continuous functions with derivatives in $L^p$ Oh if you take $L = \int_{\mathbb R} f'(t) dt$, then I think it works applying Holder. Dec12 comment Absolutely continuous functions with derivatives in $L^p$ I thought about that. Then doesn't the constant $L$ that you get depend on $x$ and $y$? Dec11 comment How to know a function is in $L^p$. Oh sorry about my confusion! Thank you! Dec11 comment How to know a function is in $L^p$. I am not sure to see why we get what we want. We don't know anything about $m(\{x | f(x)^p > t\}$. It seems to me that you have only proven the well-known identity $\int g(x) = \int m(\{x | g(x) > t\}$ for any function $g$. Dec11 comment On an identity about integrals I see. Thank you, I did not realize that! Dec10 comment On an identity about integrals Thanks! The way you put it is really intuitive! Dec10 comment On one property of the Lebesgue Measure @Nate. I am not sure I see how the answers address this question. What do you mean by taking complements? Dec10 comment On one property of the Lebesgue Measure @Jacob. $I$ is an arbitrary measurable subset of $[0,1]$.