The Unapologetic Mathematician

Mathematics for the interested outsider

Integrating Simple Functions

We start our turn from measure in the abstract to applying it to integration, and we start with simple functions. In fact, we start a bit further back than that even; the simple functions are exactly the finite linear combinations of characteristic functions, and so we’ll start there.

Given a measurable set E, there’s an obvious way to define the integral of the characteristic function \chi_E: the measure \mu(E)! In fact, if you go back to the “area under the curve” definition of the Riemann integral, this makes sense: the graph of \chi_E is a “rectangle” (possibly in many pieces) with one side a line of length 1 and the other “side” the set E. Since \mu(E) is our notion of the “size” of E, the “area” will be the product of 1 and \mu(E). And so we define


That is, the integral of the characteristic function \chi_E with respect to the measure \mu is \mu(E). Of course, this only really makes sense if \mu(E)<\infty.

Now, we’re going to want our integral to be linear, and so given a linear combination f=\sum\alpha_i\chi_{E_i} we define the integral

\displaystyle\int f\,d\mu=\int\left(\sum\alpha_i\chi_{E_i}\right)\,d\mu=\sum\alpha_i\int\chi_{E_i}\,d\mu=\sum\alpha_i\mu(E_i)

Again, this only really makes sense if all the E_i associated to nonzero \alpha_i have finite measure. When this happens, we call our function f “integrable”.

Since every simple function f is a finite linear combination of characteristic functions, we can always use this to define the integral of any simple function. But there might be a problem: what if we have two different representations of a simple function as linear combinations of characteristic functions? Do we always get the same integral?

Well, first off we can always choose an expression for f so that the E_i are disjoint. As an example, say that we write f=\alpha\chi_A+\beta\chi_B, where A and B overlap. We can rewrite this as f=\alpha\chi_{A\setminus B}+\beta\chi_{B\setminus A}+(\alpha+\beta)\chi_{A\cap B}. If f is integrable, then A and B both have finite measure, and so \mu is subtractive. Thus we can verify

\displaystyle\begin{aligned}\int\alpha\chi_{A\setminus B}+\beta\chi_{B\setminus A}+(\alpha+\beta)\chi_{A\cap B}&=\alpha\mu(A\setminus B)+\beta\mu(B\setminus A)+(\alpha+\beta)\mu(A\cap B)\\&=\alpha\left(\mu(A)-\mu(A\cap B)\right))+\beta\left(\mu(B)-\mu(A\cap B)\right)+(\alpha+\beta)\mu(A\cap B)\\&=\alpha\mu(A)+\beta\mu(B)\\&=\int\alpha\chi_A+\beta\chi_B\,d\mu\end{aligned}

Thus given any representation the corresponding disjoint representation gives us the same integral.

But what if we have two different disjoint representations f=\sum\alpha_i\chi_{E_i} and f=\sum\beta_j\chi_{F_j}? Our function can only take a finite number of nonzero values \{\gamma_k\}. We can define G_k to be the (measurable) set where f takes the value \gamma_k. For any given k, we can consider all the i so that \alpha_i=\gamma_k. The corresponding sets E_i must be a disjoint partition of G_k, and additivity tells us that the sum of these \mu(E_i) is equal to \mu(G_k). But the same goes for the F_j corresponding to values \beta_j=\gamma_k. And so both our representations give the same integral as f=\sum\gamma_k\chi_{G_k}. Everything in sight is linear, so this is all very straightforward.

At the end of the day, the integral of any simple function f is well-defined so long as all the preimage G_k of each nonzero value \gamma_k has a finite measure. Again, we call these simple functions “integrable”.


May 24, 2010 - Posted by | Analysis, Measure Theory


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    Pingback by Basic Properties of Integrable Simple Functions « The Unapologetic Mathematician | May 25, 2010 | Reply

  2. I love your blog. It’s really telling a great story (or many stories).

    But that’s a problem…I started following it about a year ago and I want to catch up on what had happened up until then. There’s no easy way to navigate through the older posts (except by tediously looking day by day in the calendar). The ‘Select Category’ list helps a bit but it doesn’t give a useful scan of the development of the story.

    Do you have an outline/table of contents of what’s happened so far? Some kind of trail of breadcrumbs to get a feel for the coarser grained subjects being followed. Or just a list of all titles forever?

    An outline of where you plan on going from here might be nice (but also might give away too much of the plot).

    Comment by Mitch | May 25, 2010 | Reply

  3. Unfortunately, I don’t really. Personally, I find the search bar helps when I’m going back to find earlier topics to link back to, but I can recognize relevant posts on the results pages because I wrote them all to begin with.

    Comment by John Armstrong | May 25, 2010 | Reply

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  9. […] the alternate approach proceeds by defining the integral of a simple function as before, and defining general integrals of non-negative functions by the supremum above. General […]

    Pingback by An Alternate Approach to Integration « The Unapologetic Mathematician | June 18, 2010 | Reply

  10. […] is not integrable, but a.e., there is really only one possibility: there is no upper bound on the integrals of simple functions smaller than . And so in this situation it makes sense to […]

    Pingback by Extending the Integral « The Unapologetic Mathematician | June 21, 2010 | Reply

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