Representing element of a class in $L^1[0,1]$ 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?
 A: 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.
A: 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.

