The Unapologetic Mathematician

Mathematics for the interested outsider

Integration with Respect to a Signed Measure

If \mu is a signed measure then we know that the total variation \lvert\mu\rvert is a measure. It then makes sense to discuss whether or not a measurable function f is integrable with respect to \lvert\mu\rvert. In this case, f will be integrable with respect to both \mu^+ and \mu^-. Indeed, since \mu^(E)\leq\lvert\mu\rvert(E) this is obviously true for simple f, and general integrable functions are limits of simple integrable functions.

This being the case, we can define both integrals

\displaystyle\begin{aligned}&\int f\,d\mu^+\\&\int f\,d\mu^-\end{aligned}

and, since \mu=\mu^+-\mu-, it makes sense to define

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

This integral shares some properties with “positive” integrals. For instance, it’s clearly linear:

\displaystyle\int \alpha f+\beta g\,d\mu=\alpha\int f\,d\mu+\beta\int g\,d\mu

Unfortunately, it doesn’t play well with order. Indeed, if E is a measurable \mu-negative set, then \chi_E\geq0 everywhere, but


This throws off most of our basic properties. However, some can be salvaged. It’s no longer necessary that f=0 a.e. for the integral to be zero, but it’s sufficient. And, thus, if f=g a.e. then their integrals are equal, although the converse doesn’t hold.

One interesting fact is that for every measurable set E we find


where we take the supremum over all measurable functions f with \lvert f(x)\rvert\leq1 everywhere. Indeed, if we take a Hahn decomposition X=A\uplus B for \mu, then since E is measurable so are E\cap A and E\cap B. If we take f^+=\chi_{E\cap A} and f^-=\chi{E\cap B}, then we find

\displaystyle\begin{aligned}\int_Ef\,d\mu&=\int_E(f^+-f^-)\,d\mu^+-\int_E(f^+-f^-)\,d\mu^-\\&=\int_Ef^+\,d\mu^+-\int_Ef^-\,d\mu^+-\int_Ef^+\,d\mu^-+\int_Ef^-\,d\mu^-\\&=\int_E\chi_{E\cap A}\,d\mu^+-\int_E\chi_{E\cap B}\,d\mu^+-\int_E\chi_{E\cap A}\,d\mu^-+\int_E\chi_{E\cap B}\,d\mu^-\\&=\int_{E\cap A}\,d\mu^+-\int_{E\cap B}\,d\mu^+-\int_{E\cap A}\,d\mu^-+\int_{E\cap B}\,d\mu^-\\&=\mu^+(E\cap A)-\mu^+(E\cap B)-\mu^-(E\cap A)+\mu^-(E\cap B)\\&=\mu^+(E\cap A)-0-0+\mu^-(E\cap B)\\&=\mu^+(E\cap A)+\mu^+(E\cap B)+\mu^-(E\cap A+\mu^-(E\cap B)\\&=\mu^+(E)+\mu^-(E)\\&=\lvert\mu\rvert(E)\end{aligned}

Thus we can actually attain this value. Can we get any larger? No. We can’t achieve anything by adding to the value of f outside E, since the integral is only taken over E anyway. And within E we could only increase the positive component of the integral by increasing the value of f in E\cap A, or increase the negative component by decreasing the value of f in E\cap B. Either way, we’d make some \lvert f(x)\rvert>1, which isn’t allows. Thus the total variation over E is indeed this supremum.

June 30, 2010 Posted by | Analysis, Measure Theory | 2 Comments