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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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    @Sam: I hope you don'$t$ mind $t$hat 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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    +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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    Quite so. Or, as written explicitly in my post, any other countable set Z.2012-01-24