The Unapologetic Mathematician

Mathematics for the interested outsider

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

   

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