Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:,,

Based on 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?

share|cite|improve this question
I have a slight correction. What you have there is technically the Darboux integral. The definition of the Riemann integral uses tagged partitions and is equivalent to the Darboux integral. – kahen Oct 13 '12 at 15:03
It is the Darboux definition of the Riemann Integral. Should I write that in the title to make things clearer? – Nameless Oct 13 '12 at 15:10
@Nameless : Riemann's integral was defined by Riemann, not by Darboux. If Darboux defined something it is Darboux's integral, whether they are equivalent or not in whatever sense. If you are using the Darboux integral you should write so. – Patrick Da Silva Oct 13 '12 at 17:18
Very well. I will update the post – Nameless Oct 13 '12 at 17:24
@PatrickDaSilva Ok any thoughts for the proof? Have I done a mistake or is the proof here incorrect – Nameless Oct 14 '12 at 9:48
up vote 1 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.