The Unapologetic Mathematician

Extending the Integral

Given an integrable function $f$, we’ve defined the indefinite integral $\nu(E)$ to be the set function

$\displaystyle\nu(E)=\int\limits_Ef\,d\mu=\int f\chi_E\,d\mu$

This is clearly real-valued, and we’ve seen that it’s countably additive. If $f$ is a.e. non-negative, then $\nu(E)$ will also be non-negative, and so the indefinite integral is a measure. Since $f$ is integrable we see that

$\displaystyle\nu(X)=\int f\,d\mu<\infty$

and so $\nu$ is a totally finite measure.

But this situation feels a bit artificially restrictive in a couple ways. First of all, measures can be extended real-valued — why do we never find $\nu(E)=\pm\infty$? Well, it makes sense to extend the definition of at least the symbol of integration a bit. If $f$ is not integrable, but $f\geq0$ a.e., there is really only one possibility: there is no upper bound on the integrals of simple functions smaller than $f$. And so in this situation it makes sense to define

$\displaystyle\int f\,d\mu=\infty$

Similarly, if $f\leq0$ a.e. and fails to be integrable, it makes sense to define

$\displaystyle\int f\,d\mu=-\infty$

In general, we can break a function $f$ into its positive and negative parts $f^+$ and $f^-$, and then define

$\displaystyle f\,d\mu=\int f^+\,d\mu-\int f^-\,d\mu$

for all functions for which at most one of $f^+$ and $f^-$ fails to be integrable. That is, if the positive part $f^+$ is integrable but the negative part $f^-$ is not, then the integral can be defined to be $-\infty$. If the negative part is integrable but the positive part isn’t, we can define the integral to be $\infty$. If both positive and negative parts are integrable then the whole function is integrable, while if neither part is integrable we still leave the integral undefined. We don’t know in general how to deal with the indeterminate form $\infty-\infty$.

And so now we find that any a.e. non-negative function — integrable or not — defines a measure by its indefinite integral. If $f$ isn’t integrable, then we get an extended real-valued set function, but this doesn’t prevent it from being a measure. As a matter of terminology, we should point out that we don’t call a function whose integral is now defined to be positive or negative $\infty$ to be “integrable”. That term is still reserved for those functions whose indefinite integrals are totally finite, as above.

June 21, 2010 - Posted by | Analysis, Measure Theory