The Unapologetic Mathematician

Iterated Integrals V

Our iterated integrals worked out nicely over $n$-dimensional intervals because these regions are simple products of one-dimensional intervals. When working over more general sets, though, it’s not so nice. Still, we can often do almost as well.

Unlike our earlier approach, though, we won’t peel off an integral from the outside first, but from an inside. That is, instead of writing

$\displaystyle\int\limits_Rf(x)\,dx=\int\limits_{a^k}^{b^k}\int\limits_{R_k}f(\hat{x},x^k)\,d\hat{x}\,dx^k$

we’ll write

$\displaystyle\int\limits_Rf(x)\,dx=\int\limits_{R_k}\int\limits_{a^k}^{b^k}f(\hat{x},x^k)\,dx^k\,d\hat{x}$

Underlying all of these equations is the assumption that each of the integrals we write down exists, or at least the inner ones don’t fail to exist often enough to cause the outer ones to also fail to exist.

Okay, so let’s say that we’ve got some bounded region $S$ contained in an $n$-dimensional interval $R$. Just like before, we project $S$ into the plane with $x^k=0$ to get the $n-1$-dimensional region $S_k$. This will be contained in the projection $R_k$ of $R$. Again, we write $\hat{x}$ for the remaining $n-1$ variables.

Here’s where our restriction comes in: assume that $S$ is contained between the graphs of two integrable functions on $S_k$. That is, we want

$\displaystyle \mathrm{Cl}(S)=\left\{\left(x^1,\dots,x^n\right)\big\vert\hat{x}\in\mathrm{Cl}(S_k),g_1(\hat{x})\leq x^k\leq g_2(\hat{x})\right\}$

where $g_1:S_k\rightarrow\mathbb{R}$ and $g_2:S_k\rightarrow\mathbb{R}$ are two integrable functions defined on $S_k$.

Now we can try to evaluate the integral

\displaystyle\begin{aligned}\int\limits_Sf\,dx&=\int\limits_Rf(x)\chi_S(x)\,dx\\&=\int\limits_{R_k}\int\limits_{a^k}^{b^k}f(x)\chi_S(x)\,dx^k\,d\hat{x}\\&=\int\limits_{S_k}\int\limits_{a^k}^{b^k}f(x)\chi_S(x)\,dx^k\,d\hat{x}\\&=\int\limits_{S_k}\int\limits_{g_1(\hat{x})}^{g_2(\hat{x})}f(x)\chi_S(x)\,dx^k\,d\hat{x}\\&=\int\limits_{S_k}\int\limits_{g_1(\hat{x})}^{g_2(\hat{x})}f(x)\,dx^k\,d\hat{x}\end{aligned}

We first replace $R_k$ by $S_k$ because $\chi_S(x)$ is identically zero wherever $\hat{x}$ is outside of $S_k$, and thus so will the whole inner integral. Then we replace the limits of integration for the inner integral by $g_1(\hat{x})$ and $g_2(\hat{x})$ because for any fixed $\hat{x}$ the function $\chi_S(x)$ is zero for $x^k$ below the former and above the latter.

If we’re lucky, $S_k$ itself can be written in a similar form, as the region between the graphs of two integrable functions in another coordinate direction, and so on. Each inner integral’s integrand depends on the remaining variables of later integrals, as before. But now the bounds of integration can also depend on later variables, which adds new complication. However, once a given variable has been integrated away, nothing further out ever depends on it again, and so this whole procedure is still well-defined.

December 31, 2009 Posted by | Analysis, Calculus | 9 Comments

Before I state today’s proposition, I need to define what I mean by saying that two sets in $\mathbb{R}^n$ are “nonoverlapping”. Intuitively, we might think that this means they have no intersection, but that’s not quite it. We’ll allow some intersection, but only at boundary points. Since the regions we’re interested in for integrals are Jordan measurable, and their boundaries have zero Jordan content, so we know changing things along these boundaries in an integral will make no difference.

Let $\left\{S_i\right\}_{i=1}^n$ be a collection of bounded regions in $\mathbb{R}^n$, so that any two of these regions are nonoverlapping. We define their union

$\displaystyle S=\bigcup\limits_{i=1}^nS_i$

and let $f:S\rightarrow\mathbb{R}$ be a bounded function defined on this union. Then $f$ is integrable on $S$ if and only if it’s integrable on each $S_i$, and we find

$\displaystyle\int\limits_Sf\,dx=\sum\limits_{i=1}^n\int\limits_{S_i}f\,dx$

Indeed, if $R$ is some $n$-dimensional interval containing $S$, then it will also contain each of the smaller regions $S_i$, and we can define

$\displaystyle\int\limits_Sf\,dx=\int\limits_Rf(x)\chi_S(x)\,dx$

and

$\displaystyle\int\limits_{S_i}f\,dx=\int\limits_Rf(x)\chi_{S_i}(x)\,dx$

We can use Lebesgue’s condition to establish our assertion about integrability. On the one hand, the discontinuities of $f$ in $S_i$ must be contained within those of $S$, so if the latter set has measure zero then so must the former. On the other hand, the discontinuities of $f$ consist of those within each of the $S_i$, and maybe some along the boundaries. Since the boundaries have measure zero, and we assume that the discontinuities in each $S_i$ are of measure zero, their (countable) union will also have measure zero. And then so must the set of discontinuities of $f$ in $S$ have measure zero as well.

The inclusion-exclusion principle tells us that we can rewrite the characteristic function of $S$:

\displaystyle\begin{aligned}\chi_S&=\chi_{S_1\cup\dots\cup S_n}\\&=\sum\limits_{1\leq i\leq n}\chi_{S_i}-\sum\limits_{1\leq i

We can put this into our integral and use the fact that integrals are additive with respect to finite sums in the integrand

\displaystyle\begin{aligned}\int\limits_Sf\,dx&=\int\limits_Rf(x)\chi_S(x)\,dx\\&=\sum\limits_{1\leq i\leq n}\int\limits_Rf(x)\chi_{S_i}(x)\,dx-\sum\limits_{1\leq i

But we assumed that all the $S_i$ are nonoverlapping, so any intersection of two or more of them must lie only along their boundaries. And since these boundaries all have Jordan content zero the integrals over them must come out to zero. We’re left with only the sum over each subregion, as we wanted.

The Mean Value Theorem for Multiple Integrals

As in the single variable case, multiple integrals satisfy a mean value property.

First of all, we should note that, like one-dimensional Riemann-Stieltjes integrals with increasing integrators, integration preserves order. That is, if $f$ and $g$ are both integrable over a Jordan-measurable set $S$, and if $f(x)\leq g(x)$ at each point $x\in S$, then we have

$\displaystyle\int\limits_Sf(x)\,dx\leq\int\limits_Sg(x)\,dx$

This is a simple consequence of the definition of a multiple integral as the limit of Riemann sums, since every Riemann sum for $f$ will be smaller than the corresponding sum for $g$.

Now if $f$ and $g$ are integrable on $S$ and $g(x)\geq0$ for every $x\in S$, then we set $m=\inf f(S)$ and $M=\sup f(S)$ — the infimum and supremum of the values attained by $f$ on $S$. I assert that there is some $\lambda$ in the interval $[m,M]$ so that

$\displaystyle\int\limits_Sf(x)g(x)\,dx=\lambda\int\limits_Sg(x)\,dx$

In particular, we can set $g(x)=1$ and find

$\displaystyle mc(S)\leq\int\limits_Sf(x)\,dx\leq Mc(S)$

giving bounds on the integral in terms of the Jordan content of $S$. Incidentally, $g(x)\,dx$ here is serving a similar role to the integrator $d\alpha$ in the integral mean value theorem for Riemann-Stieltjes integrals.

Okay, so since $g(x)\geq0$ we have $mg(x)\leq f(x)g(x)\leq Mg(x)$ for every $x\in S$. Since integration preserves order, this yields

$\displaystyle m\int\limits_Sg(x)\,dx\leq\int\limits_Sf(x)g(x)\,dx\leq M\int\limits_Sg(x)\,dx$

If the integral of $g$ is zero, then our result automatically holds for any value of $\lambda$. Otherwise we can divide through by this integral and set

$\displaystyle\lambda=\frac{\displaystyle\int\limits_Sf(x)g(x)\,dx}{\displaystyle\int\limits_Sg(x)\,dx}$

which will be between $m$ and $M$.

One particularly useful case is when $S$ has Jordan content zero. In this case, we find that any integral over $S$ is itself automatically zero.

December 29, 2009 Posted by | Analysis, Calculus | 3 Comments

Inclusion-Exclusion Again

Now that we know a little more about characteristic functions, let’s see how they can be used to understand the inclusion-exclusion principle. Our first pass was through a very categorified lens, which was a neat tie-in to Euler characteristics. But we want something a little more down-to-Earth.

So, let’s take a finite collection $\left\{S_i\right\}_{i=1}^n$ of subsets of a set $X$. To each one we have a corresponding idempotent function $\chi_{S_i}$, which is ${1}$ for points in $S_i$ and ${0}$ elsewhere. We can take the complement of a subset in the language of idempotents as usual: $1-\chi_{S_i}=\chi_{{S_i}^c}$. This function is ${1}$ for points not in $S_i$, and ${0}$ in $S_i$.

Now let’s take and multiply all these complementary functions together to get

$\displaystyle\prod\limits_{i=1}^n\left(1-\chi_{S_i}\right)=\prod\limits_{i=1}^n\chi_{{S_i}^c}=\chi_{{S_1}^c\cap\dots\cap{S_n}^c}$

By DeMorgan’s Laws, the intersection of all these complements is the complement of the union of all the $S_i$. Thus

$\displaystyle\prod\limits_{i=1}^n\left(1-\chi_{S_i}\right)=\chi_{(S_1\cup\dots\cup S_n)^c}=1-\chi_{S_1\cup\dots\cup S_n}$

Now let’s expand out the product, just like we expand out the product of any sequence of binomials. In each factor, we choose which of the ${1}$ or the $\chi_{S_i}$ to contribute. First we get the term where we always choose ${1}$, then the terms where we choose exactly one of the $\chi_{S_i}$, then the terms where we choose exactly two of them, and so on until we choose all $n$ of them. The sign of a term is negative when we choose an odd number of the $\chi_{S_i}$, and positive when we choose an even number. That is

\displaystyle\begin{aligned}\prod\limits_{i=1}^n\left(1-\chi_{S_i}\right)&=1-\sum\limits_{1\leq i\leq n}\chi_{S_i}+\sum\limits_{1\leq i

Putting this together with the earlier expression, we find

$\displaystyle\chi_{S_1\cup\dots\cup S_n}=\sum\limits_{1\leq i\leq n}\chi_{S_i}-\sum\limits_{1\leq i

For $n=2$ this gives us back our formula for the characteristic function of the intersection of two sets. For larger $n$ it gives us the inclusion-exclusion principle in terms of characteristic functions.

December 28, 2009 Posted by | Fundamentals | 1 Comment

Characteristic Functions as Idempotents

I just talked about characteristic functions as masks on other functions. Given a function $f:X\rightarrow\mathbb{R}$ and a subset $S\subseteq X$, we can mask the function $f$ to the subset $S$ by multiplying it by the characteristic function $\chi_S$. I want to talk a little more about these functions and how they relate to set theory.

First of all, it’s easy to recognize a characteristic function when we see one: they’re exactly the idempotent functions. That is, ${\chi_S}^2=\chi_S$, and if $f^2=f$ then $f$ must be $\chi_S$ for some set $S$. Indeed, given a real number $a$, we can only have $a^2=a$ if $a=0$ or $a=1$. That is, $f(x)=0$ or $f(x)=1$ for every $x$. So we can define $S$ to be the set of $x\in X$ for which $f(x)=1$, and then $f(x)=\chi_S(x)$ for every $x\in X$. Thus the idempotents in the algebra of real-valued functions on $X$ correspond exactly to the subsets of $X$.

We can define two operations on such idempotent functions to make them into a lattice. The easier to define is the meet. Given idempotents $\chi_S$ and $\chi_T$ we define the meet to be their product:

$\displaystyle\left[\chi_S\wedge\chi_T\right](x)=\chi_S(x)\chi_T(x)$

This function will take the value ${1}$ at a point $x$ if and only if both $\chi_S$ and $\chi_T$ do, so this is the characteristic function of the intersection

$\displaystyle\chi_S\wedge\chi_T=\chi_{S\cap T}$

We might hope that the join would be the sum of two idempotents, but in general this will not be another idempotent. Indeed, we can check:

$\displaystyle(\chi_S+\chi_T)^2={\chi_S}^2+{\chi_T}^2+2\chi_S\chi_T=(\chi_S+\chi_T)+2\chi_{S\cap T}$

We have a problem exactly when the corresponding sets have a nonempty intersection, which leads us to think that maybe this has something to do with the inclusion-exclusion principle. We’re “overcounting” the intersection by just adding, so let’s subtract it off to define

$\displaystyle\left[\chi_S\vee\chi_T\right](x)=\chi_S(x)+\chi_T(x)-\chi_S(x)\chi_T(x)$

We can multiply this out to check its idempotence, or we could consider its values. If $x$ is not in $T$, then $\chi_T(x)=0$, and we find $\chi_S\vee\chi_T=\chi_S$ — it takes the value ${1}$ if $x\in S$ and ${0}$ otherwise. A similar calculation holds if $x\notin S$, which leaves only the case when $x\in S\cap T$. But now $\chi_S(x)$ and $\chi_T(x)$ both take the value ${1}$, and a quick calculation shows that $\chi_S\vee\chi_T$ does as well. This establishes that

$\displaystyle\chi_S\vee\chi_T=\chi_{S\cup T}$

We can push further and make this into an orthocomplemented lattice. We define the orthocomplement of an idempotent by

$\displaystyle\left[\neg\chi_S\right](x)=1-\chi_S(x)$

This function is ${1}$ wherever $\chi_S$ is ${0}$, and vice-versa. That is, it’s the characteristic function of the complement

$\displaystyle\neg\chi_S=\chi_{X\setminus S}$

So we can take the lattice of subsets of $X$ and realize it in the nice, concrete algebra of real-valued functions on $X$. The objects of the lattice are exactly the idempotents of this algebra, and we can build the meet and join from the algebraic operations of addition and multiplication. In fact, we could turn around and do this for any commutative algebra to create a lattice, which would mimic the “lattice of subsets” of some “set”, which emerges from the algebra. This sort of trick is a key insight to quite a lot of modern geometry.

December 23, 2009 Posted by | Algebra, Fundamentals, Lattices | 2 Comments

Integrals Over More General Sets

To this point we’ve only discussed multiple integrals over $n$-dimensional intervals. But often we’re interested in more general regions, like circles or ellipsoids or even more rectangular solids that are just tilted with respect to the coordinate axes. How can we handle integrating over these more general sets?

One attempt might be to fill up the region from inside. We can chop up regions into smaller pieces, each of which is an $n$-dimensional interval. But overall this requires an incredibly involved limiting process, and we’ll never get anything calculated that way.

Instead, we come at it from outside. Given a bounded region $S\subseteq\mathbb{R}^n$, we can put it into an $n$-dimensional interval $R$. Then if $f:S\rightarrow\mathbb{R}^n$ is a function, we can try to define some sort of integral of $f$ over $S$ in terms of its integral over $R$.

The obvious problem with $\int_Rf\,dx$ is that it includes all the points in $R$ that aren’t in $S$, and we don’t want to include the integral of $f$ over that region. Worse, what if $f$ has a big cluster of discontinuities within $R$ but outside of $S$? Clearly that shouldn’t make $f$ fail to be integrable over $S$. What we need is to mask off $S$, like a stencil or masking tape does when painting.

The mask we’ll use is the characteristic function of $S$, which I’ve mentioned before. I’ll actually go more deeply into them in a bit, but for now we’ll recall that the characteristic function of a set $S$ is written $\chi_S$, and it’s defined as

$\displaystyle\chi_S(x)=\left\{\begin{matrix}1&:x\in S\\{0}&:x\notin S\end{matrix}\right.$

Now look what happens when we multiply our function by this mask:

$\displaystyle f(x)\chi_S(x)=\left\{\begin{matrix}f(x)&:x\in S\\{0}&:x\notin S\end{matrix}\right.$

Now our function has been redefined outside $S$ by setting it equal to zero there. Then we proceed to define

$\displaystyle\int\limits_Sf(x)\,dx=\int\limits_Rf(x)\chi_S(x)\,dx$

The integral over $R$ should be the “integral” over $S$ plus the “integral” over the region $R\setminus S$ outside $S$. I put these in quotes because we haven’t really defined what these integrals mean (that’s what we’re trying to do!), but we can make reasonable assertions about properties that they should have, whatever they are.

Now $f(x)\chi_S(x)$ is zero on the outer region, so the second integral is zero. And $f(x)\chi_S(x)$ is just equal to $f(x)$ inside $S$, so when “integrating” over $S$ itself we may as well just drop the $\chi_S$ factor. This justifies the definition of integrating over $S$.

It should be clear that this definition doesn’t depend on the interval $R$ at all. Indeed, if we have two different intervals $R_1$ and $R_2$ which both contain $S$, then any region the one contains which the other does not must fall outside of $S$, and $f(x)\chi_S(x)$ will be zero there anyway, and so will the difference between their integrals.

Now, this definition isn’t without its problems. The clearest of which is that we’ve almost certainly introduced new discontinuities. All around the boundary of $S$, if $f$ wasn’t already zero it suddenly and discontinuously becomes zero when we add our mask. This could cause trouble when trying to integrate the masked function over $R$. We must ask that $S$ be Jordan measurable, because this will happen if and only if the boundary $\partial S$ has zero Jordan content, and thus zero outer Lebesgue measure. Since the collection of new discontinuities must be contained in this boundary, it will also have measure zero.

This leads us to an integrability criterion over a Jordan measurable set $S$: $f$ will be integrable over $S$ if and only if the discontinuities of $f$ in $S$ form a set of measure zero. Indeed, Lebesgue’s condition tells us that the discontinuities of $f(x)\chi_S(x)$ must have measure zero. These discontinuities either come from those of $f$ inside $S$, or from the characteristic function $\chi_S$ at the boundary $\partial S$. By assuming that $S$ is Jordan measurable, we force the second kind to have measure zero, and so the total collection of discontinuities will have measure zero and satisfy Lebesgue’s condition if and only if the discontinuities of $f$ inside $S$ do.

December 22, 2009 Posted by | Analysis, Calculus | 4 Comments

Iterated Integrals IV

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 bigger?

In this case, we’re considering integrating an integrable function $f:R\rightarrow\mathbb{R}$ over some $n$-dimensional interval $R=[a^1,b^1]\times\dots\times[a^n,b^n]$. We want something like iterated integrals to allow us to evaluate this multiple integral. We’ll do this by peeling off a single integral from the outside and leaving an integral over an $n-1$-dimensional integral inside.

Specifically, we can project the interval $R$ onto the coordinate hyperplane defined by $x^k=0$ just by leaving the coordinates $x^i$ of each point $x\in R$ the same if $i\neq k$ and setting $x^k=0$. We’ll call the resulting interval

$\displaystyle R_k=[a^1,b^1]\times\dots\times\widehat{[a^k,b^k]}\times\dots\times[a^n,b^n]$

where the wide hat means that we just leave out that one factor in the product. We’ll also write $\hat{x}$ to mean the remaining coordinates on $R_k$.

Essentially, we want to integrate first over $R_k$, and then let $x^k$ run from $a^k$ to $b^k$. We have a collection of assertions that parallel those from the two-dimensional case

• $\displaystyle{\int\limits_-}_Rf(x)\,dx\leq{\int\limits_-}_{a^k}^{b^k}{\int\limits^-}_{R_k}f(\hat{x},x^k)\,d\hat{x}\,dx^k\leq{\int\limits^-}_{a^k}^{b^k}{\int\limits^-}_{R_k}f(\hat{x},x^k)\,d\hat{x}\,dx^k\leq{\int\limits^-}_Rf(x)\,dx$
• $\displaystyle{\int\limits_-}_Rf(x)\,dx\leq{\int\limits_-}_{a^k}^{b^k}{\int\limits_-}_{R_k}f(\hat{x},x^k)\,d\hat{x}\,dx^k\leq{\int\limits^-}_{a^k}^{b^k}{\int\limits_-}_{R_k}f(\hat{x},x^k)\,d\hat{x}\,dx^k\leq{\int\limits^-}_Rf(x)\,dx$
• If $\int_Rf(x)\,dx$ exists, then we have
$\displaystyle\int\limits_Rf(x)\,dx=\int\limits_{a^k}^{b^k}{\int\limits_-}_{R_k}f(\hat{x},x^k)\,d\hat{x}\,dx^k=\int\limits_{a^k}^{b^k}{\int\limits^-}_{R_k}f(\hat{x},x^k)\,d\hat{x}\,dx^k$

with a copy of these three for each index $k$ between ${1}$ and $n$. The proofs of these are pretty much identical to the proofs in the two-dimensional case, and so I’ll just skip them.

Anyhow, once we’ve picked one of the $n$ variables and split it off as the outermost integral, we’re left with an $n-1$-dimensional integral on the inside. We can pick any one of these variables and split it off, leaving an $n-2$-dimensional integral on the inside, and so on. For each of the $n!$ orderings of the original $n$ variables, we get a way of writing the $n$-dimensional integral over $R$ as a sequence of $n$ integrals, each over a one-dimensional interval. Now, we may find some of these iterated integrals easier to evaluate than others, but in principle, if each of the $m$-dimensional integrals in the sequence exists it doesn’t matter which of the orderings we use.

So, for example, if we’re considering a bounded function $f$ defined on a three-dimensional interval $R=[a^1,b^1]\times[a^2,b^2]\times[a^3,b^3]$, we can write (up to) six different iterated integrals, assuming that all the integrals in sight exist.

\displaystyle\begin{aligned}\iiint\limits_R&=\int\limits_{a^3}^{b^3}\int\limits_{a^2}^{b^2}\int\limits_{a^1}^{b^1}f(x,y,z)\,dx\,dy\,dz=\int\limits_{a^2}^{b^2}\int\limits_{a^3}^{b^3}\int\limits_{a^1}^{b^1}f(x,y,z)\,dx\,dz\,dy\\&=\int\limits_{a^3}^{b^3}\int\limits_{a^1}^{b^1}\int\limits_{a^2}^{b^2}f(x,y,z)\,dy\,dx\,dz=\int\limits_{a^1}^{b^1}\int\limits_{a^3}^{b^3}\int\limits_{a^2}^{b^2}f(x,y,z)\,dy\,dz\,dx\\&=\int\limits_{a^2}^{b^2}\int\limits_{a^1}^{b^1}\int\limits_{a^3}^{b^3}f(x,y,z)\,dz\,dx\,dy=\int\limits_{a^1}^{b^1}\int\limits_{a^2}^{b^2}\int\limits_{a^3}^{b^3}f(x,y,z)\,dz\,dy\,dx\end{aligned}

December 21, 2009 Posted by | Analysis, Calculus | 4 Comments

Iterated Integrals III

I recently heard a characterization (if someone remembers the source, please let me know) of the situation in analysis as being that there are no theorems — only conjectures that don’t yet have counterexamples. Today’s counterexample is adapted from one in Apostol’s book, but it’s far simpler than his seems to be.

We 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 introduce iterated integrals, never looking back to all those messy double and triple Riemann sums again. Unfortunately, the existence of the iterated integrals — even if both of them exist and their values agree — is not enough to guarantee that the double integral exists. Today, we will see a counterexample.

Let $S$ be the set of points $(x,y)$ in the unit square $[0,1]\times[0,1]$ so that $x=\frac{m}{q}$ and $y=\frac{n}{q}$, where $\frac{m}{q}$ and $\frac{n}{q}$ are two fractions with the same denominator, each of which are in lowest terms. That is, it contains the point $\left(\frac{1}{3},\frac{2}{3}\right)$, but not the point $\left(\frac{2}{4},\frac{3}{4}\right)$, since when we write these latter two fractions in lowest terms they are no longer over a common denominator. We will consider the characteristic function $\chi_S$, which is ${1}$ on points in $S$ and ${0}$ elsewhere in the unit square.

First, I assert that both iterated integrals exist and have the same value. That is

$\displaystyle\int\limits_0^1\int\limits_0^1\chi_S(x,y)\,dx\,dy=0=\int\limits_0^1\int\limits_0^1\chi_S(x,y)\,dy\,dx$

Indeed, the set $S$ is symmetric between the two coordinates, so we only need to evaluate one of these iterated integrals and the other one will automatically have the same value.

If $y$ is an irrational number in $[0,1]$, then there is no $x$ at all so that $(x,y)\in S$. Thus we can easily calculate the inner integral

$\displaystyle\int\limits_0^1\chi_S(x,y)\,dx=\int\limits_0^10\,dx=0$

On the other hand, if $y$ is a rational number, we can write it in lowest terms as $y=\frac{n}{q}$. Then there are only a finite number of points $x=\frac{m}{q}$ having the same denominator at all. Thus we can break the interval $[0,1]$ into a finite number of pieces, on each of which the characteristic function has the constant value zero. Thus we can calculate the inner integral

$\displaystyle\int\limits_0^1\chi_S(x,y)\,dx=\sum\limits_{i=1}^q\int\limits_\frac{i-1}{q}^\frac{i}{q}\chi_S\left(x,\frac{n}{q}\right)\,dx=\sum\limits_{i=1}^q\int\limits_\frac{i-1}{q}^\frac{i}{q}0\,dx=\sum\limits_{i=1}^q0=0$

And so we see that for any $y$ the inner integral evaluates to ${0}$. Then it’s easy to calculate the outer integral

$\displaystyle\int\limits_0^10\,dy=0$

and, as we said before, the other iterated integral also has the value ${0}$.

On the other hand, the double integral does not exist. Yes, $S$ is countable, and so it has measure zero. However, it’s also dense, which means $\chi_S$ is discontinuous everywhere in the unit square.

Saying that $S$ is dense in the square means that every neighborhood of every point of the square contains some point of $S$. Indeed, consider a point $(x,y)$ in the square and some radius $\epsilon$. Since the real numbers are Archimedean, we can pick some $N>\frac{2}{\epsilon}$, and as many people on Nick‘s Twitter experiment (remember to follow @DrMathochist!) reminded us, there are infinitely many prime numbers. Thus we can pick a prime $p>\frac{2}{\epsilon}$. Then we can round $x$ up to the next larger fraction of the form $\frac{m}{p}$, which will be in lowest terms unless $m=p$, in which case we round $x$ down to $\frac{p-1}{p}$. Similarly, we can round $y$ up (or down) to a fraction $\frac{n}{p}$ in lowest terms. This gives us a new point $\left(\frac{m}{p},\frac{n}{p}\right)\in S$, and we can calculate the distance

$\displaystyle\left\lVert\left(x-\frac{m}{p},y-\frac{n}{p}\right)\right\rVert=\sqrt{\left(x-\frac{m}{p}\right)^2+\left(y-\frac{n}{p}\right)^2}\leq\sqrt{\frac{2}{p^2}}<\frac{2}{p}<\epsilon$

So there is a point in $S$ within any radius $\epsilon$ of $(x,y)$.

But now when we try to set up the upper and lower integrals to check Riemann’s condition we find that every subinterval of any partition $P$ must contain some points in $S$ and some points not in $S$. The points within $S$ tell us that the upper sum gets a sample value of ${1}$ for each subinterval, giving a total upper sum of ${1}$. Meanwhile the points outside of $S$ tell us that the lower sum gets a sample value of ${0}$ for each subinterval, giving a total lower sum of ${0}$. Clearly Riemann’s condition fails to hold, and thus the double integral

$\displaystyle\iint\limits_{[0,1]\times[0,1]}\chi_S(x,y)\,d(x,y)$

fails to exist, despite the iterated integrals existing and agreeing.

December 18, 2009 Posted by | Analysis, Calculus | 3 Comments

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

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