# The Unapologetic Mathematician

## Iterated Integrals I

We may remember from a multivariable calculus class that we can evaluate multiple integrals by using iterated integrals. For example, if $f:R\rightarrow\mathbb{R}^{\geq0}$ is a continuous, nonnegative function on a two-dimensional rectangle $R=[a,b]\times[c,d]$ then the integral

$\displaystyle\iint\limits_Rf(x,y)\,d(x,y)$

measures the volume contained between the graph $z=f(x,y)$ of the function and the $x$$y$ plane within the rectangle. If we fix some constant $\hat{y}$ between $c$ and $d$ we can calculate the single integral

$\displaystyle\int\limits_a^bf(x,\hat{y})\,dx$

which describes the area that the plane $y=\hat{y}$ cuts out of this volume. It exists because because the integrand is continuous as a function of $x$. In such classes, we make the reasonable assumption that as we vary $\hat{y}$ this area varies continuously. This gives us a continuous function on $[c,d]$, which will then be integrable:

$\displaystyle\int\limits_c^d\left(\int\limits_a^bf(x,y)\,dx\right)\,dy$

This is an “iterated integral”, since we perform more than one integral in sequence. We usually leave out the big parens and trust in the notation to tell us when the inner integral is closed. Our handwaving argument then justifies the belief that this iterated integral is the same as the double integral above. And this is true:

$\displaystyle\iint\limits_Rf(x,y)\,d(x,y)=\int\limits_c^d\int\limits_a^bf(x,y)\,dx\,dy$

but we haven’t really proven it.

Besides, we’re interested in more general situations. What if, say, $f$ is discontinuous along the whole line $(x,\hat{y})$ for some fixed $c\leq\hat{y}\leq d$? This line can be contained in an arbitrarily thin rectangle, so it has outer Lebesgue measure zero in the rectangle $R$. If these are the only discontinuities, then $f$ is integrable on $R$, but we can’t follow the above prescription anymore, even if it were actually rigorous. We need some method of handling this sort of thing.

To this end, we have five assertions relating the upper and lower single and double integrals involving a function $f$ which is defined and bounded on the rectangle $R$ above. Unfortunately, our notation for upper and lower integrals gets a little cumbersome here, and the $\LaTeX$ support on WordPress isn’t the most elegant. Still, we soldier on and write

$\displaystyle{\int\limits_-}_a^bf(x)\,dx=\underline{I}_{[a,b]}(f)$

and similarly for upper integrals, and for lower and upper double integrals. Now, our assertions:

• $\displaystyle{\int\limits_-}Rf(x,y)\,d(x,y)\leq{\int\limits_-}_a^b{\int\limits^-}_c^df(x,y)\,dy\,dx\leq{\int\limits^-}_a^b{\int\limits^-}_c^df(x,y)\,dy\,dx\leq{\int\limits^-}Rf(x,y)\,d(x,y)$
• $\displaystyle{\int\limits_-}Rf(x,y)\,d(x,y)\leq{\int\limits_-}_a^b{\int\limits_-}_c^df(x,y)\,dy\,dx\leq{\int\limits^-}_a^b{\int\limits_-}_c^df(x,y)\,dy\,dx\leq{\int\limits^-}Rf(x,y)\,d(x,y)$
• $\displaystyle{\int\limits_-}Rf(x,y)\,d(x,y)\leq{\int\limits_-}_c^d{\int\limits^-}_a^bf(x,y)\,dx\,dy\leq{\int\limits^-}_c^d{\int\limits^-}_a^bf(x,y)\,dx\,dy\leq{\int\limits^-}Rf(x,y)\,d(x,y)$
• $\displaystyle{\int\limits_-}Rf(x,y)\,d(x,y)\leq{\int\limits_-}_c^d{\int\limits_-}_a^bf(x,y)\,dx\,dy\leq{\int\limits^-}_c^d{\int\limits_-}_a^bf(x,y)\,dx\,dy\leq{\int\limits^-}Rf(x,y)\,d(x,y)$
• If $\int_Rf(x,y)\,d(x,y)$ exists, then we have
\displaystyle\begin{aligned}\int\limits_Rf(x,y)\,d(x,y)&=\int\limits_a^b{\int\limits_-}_c^df(x,y)\,dy\,dx=\int\limits_a^b{\int\limits^-}_c^df(x,y)\,dy\,dx\\&=\int\limits_c^d{\int\limits_-}_a^bf(x,y)\,dx\,dy=\int\limits_c^d{\int\limits^-}_a^bf(x,y)\,dx\,dy\end{aligned}

Okay, as ugly as all those are, they’re what we’ll prove next time.

December 16, 2009 Posted by | Analysis, Calculus | 10 Comments