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

\displaystyle\int\chi_E\,d\mu=\mu(E)

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 | 10 Comments

   

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