Direct proof of Fubini's Theorem for the Darboux Integral over a rectangle I am looking for a direct proof of the "Fubini" theorem for the Darboux Integral.
The Theorem:
Let $I_1\subseteq \mathbb{R}^n$, $I_2\subseteq \mathbb{R}^m$ be boxes and $f:I_1\times I_2\to \mathbb{R}$ be integrable. Then the iterated integrals
\begin{equation}\int_{I_1}\left(\int_{I_2}f(x,y)\;dx\right)\;dy\text{ and }\int_{I_2}\left(\int_{I_1}f(x,y)\;dy\right)\;dx\end{equation}
exist and 
\begin{equation}\int_{I_2\times I_2}f=\int_{I_1}\left(\int_{I_2}f(x,y)\;dx\right)\;dy=\int_{I_2}\left(\int_{I_1}f(x,y)\;dy\right)\;dx\end{equation}
A ($n$-th dimensional) box $I$ is a set $I= \left\{(x_1,...,x_n)\in \mathbb{R}^n:a_i\le x_i\le b_i,\ i=1,...,n\right\}$. The integral of a bounded function $f:I\to \mathbb{R}$ is defined as follows:
If $\mathcal{P}=\left\{ \mathbf{x}\in \mathbb{R}^n :c_{i-1,j}\le x_j\le c_{i,j}\ , j=1,...,n, i=1,...,k \right\}$ 
is a partition of $I$ with subpartitions $\mathcal{P}_i=\left\{\mathbf{x}\in \mathbb{R}^n :c_{i-1,j}\le x_j\le c_{i,j}\ , j=1,...,n \right\}$ we define the upper and lower Riemann sums of $f$ as 
\begin{equation}
U_{f,\mathcal{P}}:=\sum\limits_{i=1}^k\sup_{\mathbf{x}\in \mathcal{P}_i}f(\mathbf{x})vol(\mathcal{P}_i)
\text{ and }
L_{f,\mathcal{P}}:=\sum\limits_{i=1}^k\inf_{\mathbf{x}\in \mathcal{P}_i}f(\mathbf{x})vol(\mathcal{P}_i)
\end{equation} 
where $vol(\mathcal{P}_i)=\prod_{j=1}^{n}(c_{i,j}-c_{i-1,j})$.
If the numbers \begin{equation}\int\limits_{I}^{*}f:=\inf_{\mathcal{P}}U_{f,\mathcal{P}}
\text{ and }
\int\limits_{*I}f:=\sup_{\mathcal{Q}}L_{f,\mathcal{Q}}\end{equation}
are equal we say that $f$ is Riemann Integrable and denote their common value 
with the symbol
$\int\limits_{I}f$.
As I already mentioned I am looking for a somewhat direct proof from this definition. Other proofs utilising the definition with step functions can be seen here:
http://www.tau.ac.il/~tsirel/Courses/Analysis3/lect9.pdf, http://www.cmc.edu/math/publications/aksoy/Mixed_Partials.pdf, 
http://www.owlnet.rice.edu/~fjones/chap9.pdf
Based on http://math.berkeley.edu/~wodzicki/H104.F10/Integral.pdf pg 19 here is what I have done:
Let $\mathcal{P}$ be a partition of $I_1\times I_2\subseteq \mathbb{R}^{n+m}$, 
\begin{equation}\mathcal{P}=\left\{ \mathbf{z}\in \mathbb{R}^{n+m}:c_{i-1,j}\le x_j\le c_{i,j}\text{ and }\notag\\c_{i-1,j^{\prime}}\le y_{j^{\prime}-n}\le c_{i,j^{\prime}}\ , j=1,...,n, j^{\prime}=n+1,...,n+m, i=1,...,k \right\}\end{equation}
where $\mathbf{z}=(x_1,...,x_n,y_1,...,y_m)$. Consider the partitions $\mathcal{P}_1$, $\mathcal{P}_2$ of $I_1$ and $I_2$ respectively,
\begin{gather}\mathcal{P}_1=\left\{ \mathbf{x}\in \mathbb{R}^{n}:c_{i-1,j}\le x_j\le c_{i,j}\ , j=1,...,n, i=1,...,k \right\}\text{ and }\notag\\
\mathcal{P}_2=\left\{ \mathbf{y}\in \mathbb{R}^{m}:c_{i-1,j^{\prime}}\le y_{j^{\prime}-n}\le c_{i,j^{\prime}}\ j^{\prime}=n+1,...,n+m, i=1,...,k \right\}\end{gather}
Obviously $\mathcal{P}_{1i}\times\mathcal{P}_{2i}= \mathcal{P}_i$ and $\mathcal{P}_{1}\times\mathcal{P}_{2}= \mathcal{P}$. 
Then, for $i=1,...,k$
\begin{equation}\inf_{(x,y)\in \mathcal{P}_i}f(x,y)=\inf_{x\in \mathcal{P}_{1i}}\left(\inf_{y\in \mathcal{P}_{2i}}f(x,y)\right)\end{equation}
Indeed, for arbitrary $\epsilon>0$,
\begin{gather}\exists x\in \mathcal{P}_{1i}:\inf_{x\in \mathcal{P}_{1i}}\left(\inf_{y\in \mathcal{P}_{2i}}f(x,y)\right)+\frac{\epsilon}{2}>\inf_{y\in \mathcal{P}_{2i}}f(x,y)\text{ and } 
\exists y\in \mathcal{P}_{2i}:\inf_{y\in \mathcal{P}_{2i}}f(x,y)+\frac{\epsilon}{2}>f(x,y)\Rightarrow \notag\\
\exists (x,y)\in \mathcal{P}{i}:\inf_{x\in \mathcal{P}_{1i}}\left(\inf_{y\in \mathcal{P}_{2i}}f(x,y)\right)+\epsilon>f(x,y)
\end{gather}
Therefore,
\begin{equation}
L_{f,\mathcal{P}}=\sum\limits_{i=1}^k\inf_{(x,y)\in \mathcal{P}_i}f(x,y)vol(\mathcal{P}_i)=\sum\limits_{i=1}^k\inf_{x\in \mathcal{P}_{1i}}\left(\inf_{y\in \mathcal{P}_{2i}}f(x,y)\right)vol(\mathcal{P}_{1i})vol(\mathcal{P}_{2i})
\end{equation} 
How do I proceed from there?
 A: Here is the full proof:
Lemma: Let $I_1\subseteq \mathbb{R}^n$, $I_2\subseteq \mathbb{R}^m$ be boxes and $f:I_1\times I_2\to \mathbb{R}$ be bounded. Then,
\begin{gather}\int_{I_1\times I_2*}f\le \int_{I_1*}\left(\int_{I_2*}f(x,y)\;dy\right)\;dx\le 
\left\{\begin{matrix}
\displaystyle\int_{I_1}^*\left(\displaystyle\int_{I_2*}f(x,y)\;dy\right)\;dx\\
\displaystyle\int_{I_1*}\left(\displaystyle\int_{I_2}^*f(x,y)\;dy\right)\;dx\\
\end{matrix}\right\}\le
\int_{I_1}^*\left(\int_{I_2}^*f(x,y)\;dy\right)\;dx\le \int_{I_1\times I_2}^*f
\end{gather}
Proof: Observe that any partition $\Delta$ of $I_1\times I_2\subseteq \mathbb{R}^{n+m}$ can be expressed as $\mathcal{Q}\times \mathcal{R}$ where $\mathcal{Q},\mathcal{R}$ are partitions of $I_1$ and $I_2$ respectively. In addition, if for arbitrary $i=1,...,k_1, j=1,...,k_2$ we consider the product $\mathcal{Q}_{i}\times\mathcal{R}_{j}$, then we can create a partition $\mathcal{P}$ of $I_1\times I_2$ that is finer than $\Delta$ and $\mathcal{P}_{i,j}=\mathcal{Q}_{i}\times\mathcal{R}_{j}$
Therefore,
\begin{gather}\inf_{(x,y)\in \mathcal{P}_{i,j}}f(x,y)\le f(x,y)\Rightarrow \inf_{(x,y)\in \mathcal{P}_{i,j}}f(x,y)vol(\mathcal{R}_j)=
\int_{\mathcal{R}_j*}\inf_{(x,y)\in \mathcal{P}_{i,j}}f(x,y)\;dy\le \int_{\mathcal{R}_j*}f(x,y)\; dy\Rightarrow\notag\\
\inf_{(x,y)\in \mathcal{P}_{i,j}}f(x,y)vol(\mathcal{R}_j)vol(\mathcal{Q}_i)=\int_{\mathcal{Q}_i*}\inf_{(x,y)\in \mathcal{P}_{i,j}}vol(\mathcal{R}_j)\;dx\le \int_{\mathcal{Q}_i*}\left(\int_{\mathcal{R}_j*}f(x,y)\;dy\right)\;dx\Rightarrow\notag\\
\inf_{(x,y)\in \mathcal{P}_{i,j}}f(x,y)vol(\mathcal{P}_{i,j})\le\int_{\mathcal{Q}_i*}\left(\int_{\mathcal{R}_j*}f(x,y)\;dy\right)\;dx
\end{gather}
and so
\begin{equation}
L_{f,\Delta}\le L_{f,\mathcal{P}}=\sum\limits_{i=1}^{k_1}\sum\limits_{j=1}^{k_2}\inf_{(x,y)\in \mathcal{P}_{i,j}}f(x,y)vol(\mathcal{P}_{i,j})\le \sum\limits_{i=1}^{k_1}\sum\limits_{j=1}^{k_2}\int_{\mathcal{Q}_i*}\left(\int_{\mathcal{R}_i*}f(x,y)\;dy\right)\;dx
=\int_{I_1*}\left(\int_{I_2*}f(x,y)\;dy\right)\;dx\end{equation} 
Taking the supremum with respect to all partitions $\Delta$ of $I_1\times I_2$ yields that:
\begin{equation}\int_{I_1\times I_2*}f\le \int_{I_1*}\left(\int_{I_2*}f(x,y)\;dy\right)\;dx\end{equation}
which is the first inequality of this Lemma. The middle two inequalites are obvious and the last inequality follows simillarly.
