$f$ ACL $\implies \exists \nabla f$ almost everywhere We have the following definition for an absolutely continuous function:

Definition (absolutely continuous): A function $f:I\to \mathbb{R}^n$, $I \subset \mathbb{R}$ interval, is said to be absolutely continuous if, $\forall \epsilon>0$, $\exists \delta>0$ such that,$$[x_k,y_k]\subset I, k\in \{1,\dots,n\}\text{ pairwise disjoint intervals with }\sum_{k=1}^n|y_k-x_k|<\delta$$ $$\implies\sum_{k=1}^n|f(y_k)-f(x_k)|<\epsilon.$$

And then, we have the following proposition

Proposition 1: If $f:I\longrightarrow\Bbb{R}^n$ is absolutely continuous, then $f$ is differentiable at almost everywhere in $I$.

Ok, now we give the definition of a absolutely continuous on lines (ACL) function.

Definition (ACL): Let $\Omega\subset \Bbb{R}^n$ be a domain (open connected set) and $F:\Omega\longrightarrow \Bbb{R}^m$ a function. Given a $n$-dimensional parallelepiped $P=[a_1,b_1]\times \dots \times [a_n,b_n]\subset \Omega$ and given a choice of $n-1$ fixed values $x_1\in [a_1,b_1],\dots,x_{i-1}\in[a_{i-1},b_{i-1}],x_{i+1}\in[a_{i+1},b_{i+1}],\dots,x_n\in[a_n,b_n]$, we define the function
$$\begin{array}{cccc}F^P_{x_1\dots x_{i-1}x_{i+1}\dots x_n}:&[a_i,b_i]&\longrightarrow &\Bbb{R}^m\\ &x_i&\mapsto& F(x_1,\dots,x_i,\dots,x_n)\end{array}$$ which "fix" the coordinates of $F$ at $P$, except for $x_i$. Then, we say $F$ is absolutely continuous on lines (or, simply, ACL), if, for every parallelepiped $P\in \Omega$,
$$F^P_{x_1\dots x_{i-1}x_{i+1}\dots x_n} \text{ is absolutely continuous for almost every "$n-1$-choice" as above.}$$

What I want to prove, maybe throughout Proposition 1, is

Proposition 2: If $F:\Omega\subset \Bbb{R}^n\longrightarrow \Bbb{R}^m$ is ACL, then $\nabla F$ (or, partial derivatives) exists almost everywhere on $\Omega$.

"Almost everywhere" (:D) I've searched, books say "it's easy to see", "easy to prove", but I could not do it by myself. Can some one at least give me some insight of how to do it?
I really need the case where $n=m=2$. Fixing $x$ and looking to $F^R_x(y)=F(x,y)$, I've tried to use that, at those points in which $F^R_y$ is NOT absolutely continuous is a set of measure zero, Proposition 1 holds. And, then, to look to the points where $F_x^R$ is NOT differentiable (which again has measure zero) and finally to make the union (in $x$) of all these sets, hoping that this had still measure zero. Done this, I would have taken out from $R$ a set of measure zero. Doing the same for fixed $y$'s I would have reached the result. But my conjecture has prooved to be FALSE (in the sense that this set I'm looking at might NOT TO BE MEASURABLE), as we see here.
THANK YOU, GUYS!
 A: This is trivial; requires only the definitons, Proposition 1 and Tonelli.
Suppose, just to simplify the typing, that $f:\Bbb R^2\to\Bbb R$ is ACL. Define $$f_y(x)=f(x,y).$$By definition, $f$ ACL says that almost every $f_y$ is AC, hence Proposition 1 shows that $f_y$ is differentiable almost everywhere. But the definition of $\partial f/\partial x$ says, in slightly unfortunate notation, that $$\frac{\partial f}{\partial x}(x,y)=f_y'(x).$$
So: For almost every $y$, $\partial f/\partial x$ exists at $(x,y)$ for almost every $x$. Now Tonelli's Theorem shows that $\partial f/\partial x$ exists almost everywhere.
Edit Looking back at the question I see you're concerned about measurability, which I ignored above. This can't be a real problem... Ok, here's a kludge.
Say $E$ is the set of $y$ such that $f_y$ is not AC. Define $g(x,y)=0$ for $y\in E$, $f(x,y)$ for $y\notin E$. Since $\Bbb R\times E$ has measure zero it's enough to show that $\partial g/\partial x$ exists almost everywhere.
For $\delta>0$ define $$\omega_\delta g(x,y)=\sup_{0<s<t<\delta}\left|\frac{g(x+s,y)-g(x,y)}{s}-\frac{g(x+t,y)-g(x,y)}{t}\right|.$$Since every $g_y$ is continuous it's enough to take the sup over rational $s$ and $t$; hence $\omega_\delta g$ is measurable. The argument above shows that $\omega_{1/n}g\to0$ almost everywhere, and this shows that $\partial g/\partial x$ exists almost everywhere.
