# The Unapologetic Mathematician

## 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

$\displaystyle\int\chi_E\,d\mu=\mu(E)\leq0=\int0\,d\mu$

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

$\displaystyle\lvert\mu\rvert(E)=\sup\left\lvert\int\limits_Ef\,d\mu\right\rvert$

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.