# 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

1. I been following your series of posts on measure and integrals. This is getting really interesting now, thanks for taking the time to put these posts up. Comment by ip | May 30, 2010 | Reply

2. I’m not sure why this called an indefinite integral though, considering E is a subset of X Comment by ip | May 30, 2010 | Reply

As for “indefinite integral”, the idea is just as you said yesterday: “integrating” a function $f$ gives us a set function. Once you specify which set to integrate over, you get a number. In the same way, when we consider Riemann integrals we have to specify what interval to integrate over to actually get a number.

The old “indefinite integral” is a special, restricted form of the new version, where the only variation in the region is the choice of one endpoint. Comment by John Armstrong | May 30, 2010 | Reply

4. […] Integrals and Convergence I Let’s see how the notion of an indefinite integral plays with sequences of simple functions in the […]

Pingback by Indefinite Integrals and Convergence I « The Unapologetic Mathematician | May 31, 2010 | Reply

5. I think that you want to write “measurable subsets E \subset X” in the second sentence. Great series of posts. Comment by emilelahner | May 31, 2010 | Reply

6. Thanks emile. Fixed. Comment by John Armstrong | May 31, 2010 | Reply

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13. […] The Jordan Decomposition of an Indefinite Integral So, after all our setup it shouldn’t be surprising that we take an integrable function and define its indefinite integral: […]

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14. […] Continuity I We’ve shown that indefinite integrals are absolutely continuous, but today we’re going to revise and extend this notion. But first, to review: we’ve […]

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15. […] Continuity II Now that we’ve redefined absolute continuity, we should tie it back to the original one. That definition makes precise the idea of “smallness” as being bounded in size below […]

Pingback by Absolute Continuity II « The Unapologetic Mathematician | July 2, 2010 | Reply

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18. Nice Site; I’ve learned a few things in just this read!. Re the proof of absolute continuity, what do you do if f is not bounded in E;
f may even take the value oo there, right (albeit in a subset of E of measure zero)? How would the argument c*m(E)->0 ; where c is
an upper-bund proceed then?. Comment by carl | May 11, 2013 | Reply

19. To be honest I’m not sure offhand. I’m not really an analyst by training. Comment by John Armstrong | May 11, 2013 | Reply