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Suppose that $f \in L^1[0,1]$ (where $[0,1]$ gets the standard Lebesgue measure). Consider the absolutely continuous function $F(x) = \int_0^x \ f(t) \ dt$. It is a standard result that $F$ is differentiable almost everywhere and that $f^* := F \ '$ has $f^* = f$ almost everywhere. This strikes me as a bit peculiar since F only depends on the class of $f$ in $L^1[0,1]$ so, by integrating and then differentiating, we single out a privileged element of a class in $L^1[0,1]$. I'm curious what is special about the representative $f^*$? In particular, what is wrong with the set of measure zero where $f^*$ is not defined? And how is it that we are allowed to pin down a precise value at the rest of the points?

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    I like this question. However what do you mean by "*what is wrong* with the set of measure zero where $f^\ast$ is not defined"? Why should there be anything wrong with this set? Regarding the first question: It seems to me that through this process you describe, we might obtain the "smoothest" representative of the class. E.g. if there is a continuous (or smoother) representative, then we'll get that representative.2012-01-18
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    @Sam: Maybe I should have said "what *goes* wrong on the set of measure zero where $f^*$ is not defined". It seems to have something to do with whether or not $f$ has a well-defined "average value" near a point or not.2012-01-18
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    @Sam: I hope you don't mind that I added a section to my answer in which I referenced the last sentence of your comment.2012-01-18

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To be clear, we haven't quite pinned down a specific representative in $f^*$, because $F'$ need not exist everywhere (as you know). For a motivating example, if $f(x)=\frac{1}{\sqrt{|x|}}$ when $x\neq 0$, $f(0)=0$, then $F'(0)$ doesn't exist, but $F'(x)=f(x)$ when $x\neq 0$. Or for another, let $f$ be the characteristic function of an interval $[a,b]$. Then $F'(a)$ and $F'(b)$ are undefined, but $F'(x)=f(x)$ when $x\not\in\{a,b\}$. One necessary condition for $F'$ to be defined everywhere is that $f$ is equal almost everywhere to a function satisfying the intermediate value property, by Darboux's theorem.

Whenever $F'(x)$ exists, it essentially gives the average value of $f$ in a neighborhood of $x$ as you shrink the size of that neighborhood to $x$. If that average value doesn't exist, then neither does $F'(x)$. So one way to think of the result is that is says that an integrable function is almost everywhere equal to its "local average," and redefining $f$ to be its local average everywhere it exists gives you $f^*$. Those points where $f^*$ exists are sometimes called the Lebesgue points of $f$. To quote that page, "The Lebesgue points of $f$ are thus points where $f$ does not oscillate too much, in an average sense."

Summary:

  • "What is special about the representative $f^*$"? It gives the local average of $f$. As a motivating example provided by Sam in a comment, it would single out the continuous representative of $f$ if it has one.
  • "In particular, what is wrong with the set of measure zero where $f^∗$ is not defined?" These are the non-Lebesgue points of $f$, or the points where $f$ "oscillates too much, in an average sense." (I would add that they could be points where $|f|$ goes to $\infty$.)
  • "And how is it that we are allowed to pin down a precise value at the rest of the points?" I'm not sure how to answer that; we can pin down the precise value on the complement of the set where we can't. That this turns out to be almost everywhere is the beauty of the theorem.
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    Thanks for your reply. I should have really put this in the question, but do you have any idea what the representative $f^*$ is *good for*? Does it have applications? Or is it just comforting to know it exists...2012-01-18
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    The situation where $f$ is the characteristic function of a measurable set is quite interesting by the way since this gives a way to choose a representative for each measurable set (where I consider two sets to be equivalent if their symmetric difference has measure zero).2012-01-18
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    @Mike: In most applications I would guess that you will only be concerned with the values of $f$ almost everywhere anyway, so whether you refer to $f$ or $f^*$ isn't of much importance due to the fact that in general you only know $f^*$ up to a set of measure $0$. The important thing is that $f^*=f$ a.e. But then your question might be, what are the applications of the theorem that says that $F'=f$ a.e.?2012-01-18
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    @Mike: Yes the characteristic function case is interesting, and in that case the Lebesgue points of the function that lie in the set are also called the Lebesgue density points of the set. However it is not quite true that it gives you a representative of the set. You get that up to a set of measure $0$, but what you *do* get is the set of density points of the set (where $F'=1$) and the set of density points of the complement (where $F'=0$).2012-01-18
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    +1. Nice answer. Thinking of it as an averaging process (where this is possible) seems to be exactly what's going on here.2012-01-18
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Maybe an example can help. Consider a countable subset $Z\subset[0,1]$ (for example the set of rational numbers in $[0,1]$) and a summable family $(a_z)$ of nonzero numbers $a_z$ indexed by $Z$. Consider the function $f$ defined by $$ f(x)=\sum\limits_{z}a_z\cdot[x\succ z], $$ where the sum is over every $z$ in $Z$, the bracket is Iverson notation, and each symbol $\succ$ may be $\gt$ or $\geqslant$, independently of the others. Then the function $F$ is differentiable exactly on $[0,1]\setminus Z$, since, for every $x$ in $[0,1]$, $$ F(x)=\sum\limits_{z}a_z\cdot(x-z)^+. $$ Hence, for every $x\in[0,1]\setminus Z$, $$ F'(x)=\sum\limits_{z}a_z\cdot[x\geqslant z]=\sum\limits_{z}a_z\cdot[x\gt z]=\sum\limits_{z}a_z\cdot[x\succ z]=f(x). $$ For every $x\in Z$, both one-sided derivatives $F_\ell'(x)$ and $F_r'(x)$ of $F$ at $x$ exist and they are different since $$ F_\ell'(x)=\sum\limits_{z}a_z\cdot[x\gt z], $$ and $$ F_r'(x)=\sum\limits_{z}a_z\cdot[x\geqslant z]=F_\ell'(x)+a_x\ne F_\ell'(x). $$ Thus, there is no reason to select for $f^*$ any particular function such that $f^*=f$ on $[0,1]\setminus Z$ rather than another one. In the example above, every choice of $\gt$ or $\geqslant$ at each point of $Z$ is as legitimate as the others.

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    Thanks for your reply. Taking $Z = \mathbb{Q} \cap [0,1]$ is a nice easy way get an example of a strictly increasing function $[0,1] \to \mathbb{R}$ which is discontinuous on a dense (countable) set.2012-01-24
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    Quite so. Or, as written explicitly in my post, any other countable set Z.2012-01-24