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 John Armstrong | Analysis, Calculus

10 Comments »

Not ugly to me. In fact, more beautiful following your careful platform for understanding Calculus. I like your rigor, combined with your clarity of exposition. This reminds me of my joy in taking calsulus at Caltech from Tom Apostol. And keeping in touch with him since then (i.e. since 1968). His textbooks sell well, so yours should also, once you get co-authors to write them.

Comment by Jonathan Vos Post | December 16, 2009 | Reply
Jonathan, I mean the typesetting, not the content. The typesetting is terrible, which is why I tried using the $\underline{I}$ notation instead of underlining the integral sign like Apostol does.

Speaking of whom, if you’re still in contact with him could you put me in contact? I have a huge question about the exercise that inspired this coming Friday’s post.

Comment by John Armstrong | December 16, 2009 | Reply
[…] Integrals II Let’s get to proving the assertions we made last time, starting […]

Pingback by Iterated Integrals II « The Unapologetic Mathematician | December 17, 2009 | Reply
This can be found online, so I’m not betraying privacy. Just tell him that I sent you when you email him, on the chance that it might help slightly. I also attended a book-signing by his wife within the past year.

# Tom Apostol (apostol at caltech.edu)
376 Sloan
Caltech Undergraduate
1200 E. California Blvd
Pasadena CA 91125
(626) 395-4363
Project MATHEMATICS!
(626) 395-3759

Comment by Jonathan Vos Post | December 18, 2009 | Reply
[…] might guess that we can always evaluate double integrals by iterated integrals as we’ve been discussing. After all, that’s exactly what we do in multivariable calculus courses as soon as we […]

Pingback by Iterated Integrals III « The Unapologetic Mathematician | December 18, 2009 | Reply
[…] So we’ve established that as long as a double integral exists, we can use an iterated integral to evaluate it. What happens when the dimension of our space is even […]

Pingback by Iterated Integrals IV « The Unapologetic Mathematician | December 21, 2009 | Reply
I ran into Tom Apostol at Caltech this afternoon, and mentioned to him that I’d given him your email, and why. He said that he’d answer an email, so long as it wasn’t crackpot. I reassured him about you, and wished him and his wife happy holidays. So — have you emailed him yet about this rigorous instruction on Iterated Integrals?

Comment by Jonathan Vos Post | December 23, 2009 | Reply
Just doing it now, JVP.

Comment by John Armstrong | December 23, 2009 | Reply
[…] Integrals V Our iterated integrals worked out nicely over -dimensional intervals because these regions are simple products of […]

Pingback by Iterated Integrals V « The Unapologetic Mathematician | December 31, 2009 | Reply
[…] the Riemann-Stieltjes integral. First up is a question that seems natural from the perspective of iterated integrals: what can be said about the continuity of the inner […]

Pingback by Continuity of Partial Integrals « The Unapologetic Mathematician | January 12, 2010 | Reply

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