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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?

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    @PatrickDaSilva Ok any thoughts for the proof? Have I done a mistake or is the proof here http://math.berkeley.edu/~wodzicki/H104.F10/Integral.pdf incorrect2012-10-14

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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.