Lebesgue's Theorem on Differentiation of measures? I'm currently going through Ahlfors's "Lectures of Quasiconformal Mappings," and there is a part that I would like more a bit more detail on. 
Let $f$ be a orientation preserving homeomorphism on the plane, and we define a measure $A(E)$ as the area of $f(E)$. On page 18, it says that 
"This defines a locally finite additive measure, and according to a theorem of Lebesgue such a measure has a symmetric derivative a.e., that is,
$$J(z) = \lim \frac{A(Q)}{m(Q)}$$
when $Q$ is a square of center $z$ whose side tends to zero. Moreover,
$$\int_E J(z) dxdy \le A(E)$$ (we cannot yet guarantee equality)."
I think this statement is related to the Lebesgue differentiation theorem, but I haven't had luck finding actual the statement of the theorem Ahlfors is supposedly quoting. Could someone point me to a reference?
Thank you!
 A: The theorem in question is proved in Folland's "Real Analysis", actually in greater generalty than the one you seek:

Let $\nu$ be a regular signed or complex Borel measure on
  $\mathbb{R}^n$, and let $d \nu = d \lambda +f dm$ be its
  Lebesgue-Radon-Nikodym representation. Then for $m$-almost every $x \in \mathbb{R}^n$:
$\lim_{r \to 0} \frac{\nu (E_r)}{m(E_r)} = f(x)$
for every family $\{E_r\}_{r>0}$ that shrinks nicely to $x$.

Remarks:


*

*By Folland's definition, a family $\{E_r\}_{r>0}$ of subsets of $\mathbb{R}^n$ shrinks nicely to $x$ if $E_r \subset B(x,r)$ for all $r>0$ and $m(E_r) > \alpha m(B(x,r))$ for all $r>0$, for some universal constant $\alpha$.

*A Borel measure on $\mathbb{R}^n$ is regular if and only if it is locally finite, as Folland points out.

*The Lebesgue-Radon-Nikodym representation expresses $\nu$ uniquely as the sum of an absolutely continuous measure ($f dm$) and a singular measure $\lambda$. The above theorem actually shows that $f$ is uniquely determined (and gives a formula with which to compute it). If $\lambda = 0$ (i.e. $\nu$ is absolutely continuous), then one usually writes $f = \frac{d \nu}{dm}$.

*"Usual" (=positive) measures are particular instances of signed measures.


As for the inequality you mention, this is related to another theorem related to the Lebesgue-Radon-Nikodym representation. It says that if $\nu$ and $\mu$ are $\sigma$-finite measures and $\nu$ is absolutely continuous w.r.t. $\mu$, then for every $\nu$-integrable function $g$, $g (\frac{d \nu}{d \mu})$ is $\mu$-integrable and the following formula holds:
$\int g d \nu = \int g \frac{d \nu}{d \mu} d \mu$
In your case $\mu$ is Lebesgue measure and $g= \chi_E$, except that $\nu$ is not absolutely continuous w.r.t. to $m$ in general. This is the reason for the inequality: $\int g d \nu \ge \int g f dm$. Now $f dm$ is absolutely continuous w.r.t. $m$ (basically by definition) so the above formula can be applied to it. In this case $\frac{f dm}{dm} = f$ (in the sense I mention in (3)) so you get the desired inequality. If in your case the measure $\nu$ (or $A$ in your notation) has no singular part then we have equality.
Both theorems are proven in chapter 3 of Folland's book.
