The Unapologetic Mathematician

Mathematics for the interested outsider

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

   

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