I am encountering some issues in proving the following lemma 12.47 from the book "A first course on Sobolev spaces" by Giovanni Leoni.
$\mathbf{Lemma \, 12.47:}$
Take $u \in W^{m,p}(Q_{r}) \cap C^{\infty}(Q_{r})$, where $Q_{r}$ is a cube of $\mathbb{R}^{N}$ with sides parallel to the axes of length r and $m*p>N.$
Show that
$$
\vert u(x)-u(y)\vert \leq C\cdot \Big( \sum_{k=0}^{m-1}r^{k-\frac{N}{p}}\cdot \Vert \nabla^{k} u \Vert_{L^{p}(Q_{r})} + r^{m-\frac{N}{p}} \cdot \Vert D^{m}u \Vert_{L^{p}(Q_r)} \Big).
$$
I will skip straight to the passage that I do not get, I have the following integral
$$
\int_{0}^{1}\int_{t\cdot x+ Q_{(1-t)\cdot r}} \Vert \nabla^{m}u(z) \Vert \cdot \Vert z-x \Vert^{m} \cdot (1-t)^{-N-1},
$$
where $x \in Q_{r}.$ The integral is bounded from above by this other integral
$$
\int_{Q_{r}}\int_{0}^{1- \frac{\Vert z-x\Vert}{\sqrt{N}r}} \Vert \nabla^{m}u(z) \Vert \cdot \Vert z-x \Vert^{m} \cdot (1-t)^{-N-1},
$$
the author justifies this bound only with Tonelli's theorem, but I do not understand how it is applied. How do you bound $t$ from above?
If you impose that $z-x \in Q_{(1-t)\cdot r}$ then you obtain
$$
\Vert z- x\Vert \leq r \cdot (1-t)\cdot \sqrt{N} \iff t < 1- \frac{\Vert z-x\Vert}{\sqrt{N}r},
$$
but how does this relate to the problem?
If $z\in tx+Q_{(1-t)r}$, then for every $i=1,\ldots, N$, $$|z_i-x_i|\le |z_i-tx_i|+|(1-t)x_i|<(1-t)r/2+(1-t)r/2=(1-t)r.$$ Hence, $\|z-x\|\le\sqrt{N}(1-t)r$. On the other hand, since $x\in Q_r$, you have that $tx+Q_{(1-t)r}\subset Q_r$. So you use Tonelli's theorem and flip the two integrals. Does this answer your question?