# The Unapologetic Mathematician

## Iterated Integrals II

Let’s get to proving the assertions we made last time, starting with

$\displaystyle{\int\limits_-}Rf(x,y)\,d(x,y)\leq{\int\limits_-}_a^b{\int\limits^-}_c^df(x,y)\,dy\,dx\leq{\int\limits^-}_a^b{\int\limits^-}_c^df(x,y)\,dy\,dx\leq{\int\limits^-}Rf(x,y)\,d(x,y)$

where $f$ is a bounded function defined on the rectangle $R=[a,b]\times[c,d]$.

We can start by defining

$\displaystyle F(x)={\int\limits^-}_c^df(x,y)\,dy$

And we easily see that $\lvert F(x)\rvert\leq M(d-c)$, where $M$ is the supremum of $\lvert f\rvert$ on the rectangle $R$, so this is a bounded function as well. Thus the upper integral

$\displaystyle\overline{I}={\int\limits^-}_a^bF(x)\,dx={\int\limits^-}_a^b{\int\limits^-}_c^df(x,y)\,dy\,dx$

and the lower integral

$\displaystyle\underline{I}={\int\limits_-}_a^bF(x)\,dx={\int\limits_-}_a^b{\int\limits^-}_c^df(x,y)\,dy\,dx$

are both well-defined.

Now if $P_x=\{x_0,\dots,x_m\}$ is a partition of $[a,b]$, and $P_y=\{y_0,\dots,y_n\}$ is a partition of $[c,d]$, then $P=P_x\times P_y$ is a partition of $R$ into $mn$ subrectangles $R_{ij}$. We will define

\displaystyle\begin{aligned}\overline{I}_{ij}&={\int\limits^-}_{x_{i-1}}^{x_i}{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy\,dx\\\underline{I}_{ij}&={\int\limits_-}_{x_{i-1}}^{x_i}{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy\,dx\end{aligned}

Clearly, we have

$\displaystyle{\int\limits^-}_c^df(x,y)\,dy=\sum\limits_{j=1}^n{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy$

and so we find

\displaystyle\begin{aligned}{\int\limits^-}_a^b{\int\limits^-}_c^df(x,y)\,dy\,dx&={\int\limits^-}_a^b\sum\limits_{j=1}^n{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy\,dx\\&\leq\sum\limits_{j=1}^n{\int\limits^-}_a^b{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy\,dx\\&=\sum\limits_{j=1}^n\sum\limits_{i=1}^m{\int\limits^-}_{x_{i-1}}^{x_i}{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy\,dx\end{aligned}

That is

$\displaystyle\overline{I}\leq\sum\limits_{j=1}^n\sum\limits_{i=1}^m\overline{I}_{ij}$

and, similarly

$\displaystyle\underline{I}\geq\sum\limits_{j=1}^n\sum\limits_{i=1}^m\underline{I}_{ij}$

We also define $m_{ij}$ and $M_{ij}$ to be the infimum and supremum of $f$ over the rectangle $R_{ij}$, which gives us the inequalities

$\displaystyle m_{ij}(y_j-y_{j-1})\leq{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy\leq M_{ij}(y_j-y_{j-1})$

and from here we find

$\displaystyle m_{ij}\mathrm{vol}(R_{ij})\leq{\int\limits_-}_{x_{i-1}}^{x_i}{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy\,dx\leq{\int\limits^-}_{x_{i-1}}^{x_i}{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy\,dx\leq M_{ij}\mathrm{vol}(R_{ij})$

Summing on both $i$ and $j$, and sing the above inequalities, we get

$\displaystyle L_P(f)\leq\underline{I}\leq\overline{I}\leq U_P(f)$

and since this holds for all partitions $P$, the assertion that we’re trying to prove follows.

The second assertion from last time can be proven similarly, just replacing $F(x)$ by the lower integral over $[c,d]$. And then the third and fourth assertions are just the same, but interchanging the roles of $[a,b]$ and $[c,d]$. Finally, the last assertion is a consequence of the first four. Indeed, if the integral over $R$ exists, then the upper and lower integrals are equal, which collapses all of the inequalities into equalities.

December 17, 2009 Posted by | Analysis, Calculus | 1 Comment