The Unapologetic Mathematician

Mathematics for the interested outsider

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

   

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