The Unapologetic Mathematician

Mathematics for the interested outsider

Sets Measurable by an Outer Measure I

An outer measure \mu^* on a hereditary \sigma-ring \mathcal{H} is nice and all, but it’s not what we really want, which is a measure. In particular, it’s subadditive rather than additive. We want to fix this by restricting to a nice collection of sets within \mathcal{H}.

Every set E splits every other set into two pieces: the part that’s in E and the part that’s not. What we want to focus on are the sets that split every other set additively. That is, sets E\in\mathcal{H} so that for every set A\in\mathcal{H} we have

\displaystyle\mu^*(A)=\mu^*(A\cap E)+\mu^*(A\cap E^c)

We call such sets “\mu^*-measurable”. Actually, to show that E is \mu^*-measurable we just need to show that \mu^*(A)\geq\mu^*(A\cap E)+\mu^*(A\cap E^c) for every A\in\mathcal{H}, because the opposite inequality follows from the subadditivity of \mu^*.

This condition seems sort of contrived at first, and there’s really not much to justify it at first besides the foregoing handwaving. But we will soon see that this definition turns out to be useful. For one thing, the collection \overline{\mathcal{S}}\subseteq\mathcal{H} of \mu^*-measurable sets is a ring!

The proof of this fact is straightforward, but it feels like pulling a rabbit out of a hat, so follow closely. Given sets \mu^*-measurable sets E and F, we need to show that their union E\cup F and difference E\setminus F=E\cap F^c are both \mu^*-measurable as well. Saying that E is \mu^*-measurable means that for every A\in\mathcal{H} we have

\displaystyle\mu^*(A)=\mu^*(A\cap E)+\mu^*(A\cap E^c)

Saying that F is \mu^*-measurable means that for every A\in\mathcal{H} we have

\displaystyle\begin{aligned}\mu^*(A\cap E)&=\mu^*(A\cap E\cap F)+\mu^*(A\cap E\cap F^c)\\\mu^*(A\cap E^c)&=\mu^*(A\cap E^c\cap F)+\mu^*(A\cap E^c\cap F^c)\end{aligned}

We can take each of these and plug them into the first equation to find the key equation

\displaystyle\begin{aligned}\mu^*(A)&=\mu^*(A\cap E\cap F)+\mu^*(A\cap E\cap F^c)\\&+\mu^*(A\cap E^c\cap F)+\mu^*(A\cap E^c\cap F^c)\end{aligned}

Now this key equation works for A\cap(E\cup F) as well as A. We know that (E\cup F)\cap E=E and (E\cup F)\cap F=F, but (E\cup F)\cap E^c\cap F^c=\emptyset. So, sticking A\cap(E\cup F) into the key equation we find

\displaystyle\begin{aligned}\mu^*(A\cap(E\cup F))&=\mu^*(A\cap(E\cup F)\cap E\cap F)+\mu^*(A\cap(E\cup F)\cap E\cap F^c)\\&+\mu^*(A\cap(E\cup F)\cap E^c\cap F)+\mu^*(A\cap(E\cup F)\cap E^c\cap F^c)\\&=\mu^*(A\cap E\cap F)+\mu^*(A\cap E\cap F^c)+\mu^*(A\cap E^c\cap F)\end{aligned}

But the three terms on the right are the first three terms in the key equation. And so we can replace them and write

\displaystyle\begin{aligned}\mu^*(A)&=\mu^*(A\cap(E\cup F))+\mu^*(A\cap E^c\cap F^c)\\&=\mu^*(A\cap(E\cup F))+\mu^*(A\cap(E\cup F)^c)\end{aligned}

which establishes that E\cup F is \mu^*-measurable! Behold, the rabbit!

Let’s see if we can do it again. This time, we take A\cap(E\setminus F)^c=A\cap(E^c\cup F) and stick it into the key equation. We find

\displaystyle\begin{aligned}\mu^*(A\cap(E\setminus F)^c)&=\mu^*(A\cap(E^c\cup F)\cap E\cap F)+\mu^*(A\cap(E^c\cup F)\cap E\cap F^c)\\&+\mu^*(A\cap(E^c\cup F)\cap E^c\cap F)+\mu^*(A\cap(E^c\cup F)\cap E^c\cap F^c)\\&=\mu^*(A\cap E\cap F)+\mu^*(A\cap E^c\cap F)+\mu^*(A\cap E^c\cap F^c)\end{aligned}

Again we can find the three terms on the right of this equation on the right side of the key equation as well. Replacing them in the key equation, we find

\displaystyle\begin{aligned}\mu^*(A)&=\mu^*(A\cap(E\setminus F)^c)+\mu^*(A\cap(E\setminus F)^c)\end{aligned}

which establishes that E\setminus F is \mu^*-measurable as well!

March 29, 2010 - Posted by | Analysis, Measure Theory


  1. That definition is sometimes called “Caratheodory’s condition”.

    Comment by mattiast | March 29, 2010 | Reply

  2. It’s the approach used by Royden in his Real Analysis. Kind of weird, but powerful.

    Comment by Zeno | March 30, 2010 | Reply

  3. Indeed, that’s where I first saw it, Zeno. For the moment, though, I’m lifting it from Halmos. Secretly, this is my excuse to read through Halmos, digest, and represent the material (with an algebraist’s eye) in this venue.

    Comment by John Armstrong | March 30, 2010 | Reply

  4. […] Measurable by an Outer Measure II Yesterday, we showed that — given an outer measure on a hereditary -ring — the collection of -measurable […]

    Pingback by Sets Measurable by an Outer Measure II « The Unapologetic Mathematician | March 30, 2010 | Reply

  5. […] -ring . And then we can restrict this outer measure to get an actual measure on the -ring of -measurable sets. And so we ask: how does the measure relate to the measure […]

    Pingback by Measures and Induced Measures « The Unapologetic Mathematician | March 31, 2010 | Reply

  6. If you want to get rid of the magic trick, you could always just draw a venn diagram. You’re only manipulating 3 sets, so it wouldn’t be terribly cluttered at all. Then the proof would just come down to tagging which of the subsets of the whole union are mu-* measurable

    Comment by Gilbert Bernstein | March 31, 2010 | Reply

  7. The “magic” is more the way of plugging some equations back into each other in unexpected ways, until the definition we want pops out.

    Comment by John Armstrong | March 31, 2010 | Reply

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  11. […] sets” of the measurable space. This is not to insinuate that is the collection of sets measurable by some outer measure , nor even that we can define a nontrivial measure on in the first place. […]

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  12. Thankx

    Comment by Nazia irshad | May 27, 2012 | Reply

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