# The Unapologetic Mathematician

## Indefinite Integrals

We can use integrals to make a set function $\nu$ out of any integrable function $f$.

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

We call this set function the “indefinite integral” of $f$, and it is defined for all measurable subsets $E\subseteq X$. This isn’t quite the same indefinite integral that we’ve worked with before. In that case we only considered functions $f:\mathbb{R}\to\mathbb{R}$, picked a base-point $a$, and defined a new function $F$ on the domain. In our new language, we’d write $F(x)=\nu\left([a,x]\right)$, so the two concepts are related but they’re not quite the same.

Anyhow, the indefinite integral $\nu$ is “absolutely continuous”. That is, for every $\epsilon>0$ there is a $\delta$ so that $\lvert\nu(E)\rvert<\epsilon$ for all measurable $E$ with $\mu(E)<\delta$. Indeed, if $c$ is an upper bound for $\lvert f\rvert$, then we can show that

$\displaystyle\lvert\nu(E)\rvert=\left\lvert\int\limits_Ef\,d\mu\right\rvert\leq\int\limits_E\lvert f\rvert\,d\mu\leq c\mu(E)$

And so if we make $\mu(E)$ small enough we can keep $\lvert\nu(E)\rvert$ small.

Further, an indefinite integral $\nu$ is countably additive. Indeed, if $f=\chi_S$ is a characteristic function then countable additivity of $\nu$ follows immediately from countable additivity of $\mu$. And countable additivity for general simple functions $f$ is straightforward by writing each such function as a finite linear combination of characteristic functions.

With the exception of this last step, nothing we’ve said today depends on the function $f$ being simple, and so once we generalize our basic linearity and order properties we will immediately have absolutely continuous indefinite integrals.

May 27, 2010 Posted by | Analysis, Measure Theory | 19 Comments