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seen Jan 11 '13 at 21:34

Dec
15
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!
Dec
15
comment Absolutely continuous functions with derivatives in $L^p$
@Jonas. Yes thank you!
Dec
15
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.
Dec
15
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...
Dec
14
comment Absolutely continuous functions with derivatives in $L^p$
@Jonas. Is it important in this case?
Dec
14
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.
Dec
12
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.
Dec
12
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 $y$?
Dec
11
comment How to know a function is in $L^p$.
Oh sorry about my confusion! Thank you!
Dec
11
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 $g$.
Dec
11
comment On an identity about integrals
I see. Thank you, I did not realize that!
Dec
10
comment On an identity about integrals
Thanks! The way you put it is really intuitive!
Dec
10
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?
Dec
10
comment On one property of the Lebesgue Measure
@Jacob. $I$ is an arbitrary measurable subset of $[0,1]$.