On the use of Fubini's theorem in the proof of Rademacher's theorem I am reading Lectures on Lipschitz Analysis by Heinonen, where the proof of Rademacher's theorem is presented in Chapter 3.
Let $\, f:\mathbb{R}^n \to \mathbb{R}$ be Lipschitz.
In the first step of the proof, it is shown that the directional derivative of $f$ in direction $v \in \mathbb{R}^n$ exists for almost all $x \in \mathbb{R}^n$.
For any fixed $x,v \in \mathbb{R}^n$ the author defines a real-valued function $$ f_{x,v}(t)=f(x+tv)$$ which is Lipschitz, and hence differentiable at almost every $t$. Then, it is claimed that after fixing $v$, we can conclude from Fubini's theorem that $$ \lim_{t \to 0} \frac{f(x+tv) - f(x)}{t}$$ exists for almost every $x \in \mathbb{R}^n$.
Question: Why this conclusion follows from Fubini?
 A: I am posting a complete answer, based on John Dawkins answers:
Proposition:
Let $\, f:\mathbb{R}^n \to \mathbb{R}$ be Lipschitz, and let $v \in \mathbb{R}^n$. Then, the directional derivative of $f$ in direction $v$ exists for almost all $x \in \mathbb{R}^n$.
Proof:
For any fixed $x,v \in \mathbb{R}^n$ we define a real-valued function $$ f_{x,v}(t)=f(x+tv)$$ which is Lipschitz, and hence differentiable at almost every $t \in \mathbb{R}$.
We define the sets
$ B:=\{(x,t) \in \mathbb{R}^n \times \mathbb{R} | \, \, s\mapsto f(x+s\nu) $ fails to be differentiable at $ t\}  \subseteq \mathbb{R}^{n+1}.  $,
$B_x=\{ t \in \mathbb{R} |  \, \, (x,t) \in B \} \subseteq \mathbb{R}$ and
$B^t= \{x\in\Bbb R^n: (x,t)\in B\} \subseteq \mathbb{R}^n$. 
By the obesrvation above, we know that $\lambda_1(B_x)=0$ for every $x \in \mathbb{R}^n$.
Thus, by Tonelli's theorem (aplied for the function $1_B:\mathbb{R}^n \times \mathbb{R} \to \{ 0,1\}$) we get that
$$ \lambda_n\otimes\lambda_1(B) =\int_{\mathbb{R}^n  \times \mathbb{R}} 1_B= \int_{\mathbb{R}^n} ( \int_{\mathbb{R}} 1_B(x,t) dt \, ) dx = \int_{\mathbb{R}^n} ( \int_{\mathbb{R}} 1_{B_x}(t) dt \, ) dx=\int_{\Bbb R^n}\lambda_1(B_x)\,dx=0$$
Consequently (Tonelli again)
$$
0=\lambda_n\otimes\lambda_1(B)=\int_{\mathbb{R}^n  \times \mathbb{R}} 1_B= \int_{\mathbb{R}} ( \int_{\mathbb{R}^n} 1_B(x,t) dx \, ) dt = \int_{\mathbb{R}} ( \int_{\mathbb{R}^n} 1_{B^t}(x) dx \, ) dt=\int_{\Bbb R}\lambda_n(B^t) dt.
$$
Thus $\lambda_n(B^t)=0$ for a.e. $t\in\Bbb R$.
In particular, $\lambda_n(B^a)=0$ for some $a \in \mathbb{R}$. 
$$
\mathbb{R}^n \setminus B^a= \{ x \in \mathbb{R}^n | \lim_{s\to 0}{f(x+(a+s)\nu)-f(x+a\nu)\over s} \, \text{ exists} \}=\{ x \in \mathbb{R}^n | \lim_{s\to 0}{f( x+a\nu +s\nu)-f(x+a\nu)\over s} \, \text{ exists} \}
$$
Define $h:\mathbb{R}^n \to \mathbb{R}^n$ by $h(x)=x+av$. $h$ is surjective, and we showed that
 $$ \mathbb{R}^n \setminus B^a= \{ x \in \mathbb{R}^n | \lim_{t \to 0} \frac{f(h(x)+tv) - f(h(x))}{t}  \, \text{ exists}\} ,$$ so 
$$ \{ x \in \mathbb{R}^n | \lim_{t \to 0} \frac{f(x+tv) - f(x)}{t}  \, \text{ does not exists}\} \subseteq h(B^a)$$
Since $\lambda_n(h(B^a)) = \lambda_n(B^a)=0$ the proof is completed.
A: Consider the set $B:=\{(x,t): s\mapsto f(x+s\nu)$ fails to be differentiable at $t\}$. As you have noted, each of the sections $B_x$ is Lebesgue null, hence so is $B$, by Fubini:
$$
\lambda_n\otimes\lambda_1(B)=\int_{\Bbb R^n}\lambda_1(B_x)\,\lambda_n(dx)=0.
$$
Consequently (Fubini again)
$$
0=\lambda_n\otimes\lambda_1(B)=\int_{\Bbb R}\lambda_n(B^t)\,\lambda_1(dt),
$$
where $B^t$ is the "horizontal" section $\{x\in\Bbb R^n: (x,t)\in B\}$. Thus $\lambda_n(B^t)=0$ for a.e. $t\in\Bbb R$.
A: You're right that Heinonen makes a bit of a leap here. From what I wrote before we have that for a.e. pair $(x,t)$, the limit 
$$
\lim_{s\to t}{f(x+s\nu)-f(x+t\nu)\over s-t}
$$
exists. After a change of lettering, this amounts to the existence of
$$
\lim_{\epsilon\to 0}{f(x+(t+\epsilon)\nu)-f(x+t\nu)\over \epsilon},
$$
or to the existence of the limit
$$
\lim_{\epsilon\to 0}{f(\tilde x+\epsilon\nu)-f(\tilde x)\over \epsilon},
$$
where $\tilde x:=x+t\nu$. But the image of the $(n+1)$-dimensional Lebesgue null set $B$ under the linear  mapping $(x,t)\to x+t\nu$ is a null set for $\lambda_n$. 
