The Unapologetic Mathematician

Mathematics for the interested outsider

Measures

From this point in, I will define a “set function” as a function \mu whose domain is some collection of subsets \mathcal{E}\subseteq P(X). It’s important to note here that \mu is not defined on points of the set X, but on subsets of X. For some reason, a lot of people find that confusing at first.

We’re primarily concerned with set functions which take their values in the “extended real numbers” \overline{\mathbb{R}}. That is, the value of \mu(E) is either a real number, or +\infty, or -\infty, with the latter two being greater than all real numbers and less than all real numbers, respectively.

We say that such a set function \mu:\mathcal{E}\to\overline{\mathbb{R}} is “additive” if whenever we have disjoint sets E_1 and E_2 in \mathcal{E} with disjoint union E_1\uplus E_2 also in \mathcal{E}, then we have

\displaystyle\mu(E_1\uplus E_2)=\mu(E_1)+\mu(E_2)

Similarly, we say that \mu is finitely additive if for every finite, pairwise disjoint collection \{E_1,\dots,E_2\}\subseteq\mathcal{E} whose union is also in \mathcal{E} we have

\displaystyle\mu\left(\biguplus\limits_{i=1}^nE_i\right)=\sum\limits_{i=1}^n\mu(E_i)

And we say that \mu is countably additive of for every pairwise-disjoint sequence \{E_i\}_{i=1}^\infty of sets in \mathcal{E} whose union is also in \mathcal{E}, we have

\displaystyle\mu\left(\biguplus\limits_{i=1}^\infty E_i\right)=\sum\limits_{i=1}^\infty\mu(E_i)

Now we can define a “measure” as an extended real-valued, non-negative, countably additive set function \mu defined on an algebra \mathcal{A}, and satisfying \mu(\emptyset)=0. With this last assumption, we can show that a measure is also finitely additive. Indeed, given a collection \{E_1,\dots,E_n\}, just define E_i=\emptyset for i>n to get a sequence. Then we find

\displaystyle\mu\left(\biguplus\limits_{i=1}^nE_i\right)=\mu\left(\biguplus\limits_{i=1}^\infty E_i\right)=\sum\limits_{i=1}^\infty\mu(E_i)=\sum\limits_{i=1}^n\mu(E_i)

If \mu is a measure on \mathcal{A}, we say a set A\in\mathcal{A} has finite measure if \mu(A)<\infty. We say that A has “\sigma-finite” measure if there is a sequence of sets \{A_i\}_{i=1}^\infty of finite measure (\mu(A_i)<\infty) so that A\subseteq\bigcup_{i=1}^\infty A_i. If every set in A has finite (or \sigma-finite) measure, we say that \mu is finite (or \sigma-finite) on A.

Finally, we say that a measure is “complete” if for every set A of measure zero, \mathcal{A} also contains all subsets of A. That is, if A\in\mathcal{A}, \mu(A)=0, and B\subseteq A, then B\in\mathcal{A}. At first, this might seem to be more a condition on the algebra \mathcal{A} than on the measure \mu, but it really isn’t. It says that to be complete, a measure can only assign {0} to a set if all of its subsets are also in \mathcal{A}.

About these ads

March 19, 2010 - Posted by | Analysis, Measure Theory

17 Comments »

  1. This stuff is SO important, and so rarely done well. I’m confident that you can make it delightful.

    Comment by Jonathan Vos Post | March 20, 2010 | Reply

  2. thanks for these excellent posts, i been trying to teach my self measure theory i’m finding these posts very useful.
    looking forward to the rest

    Comment by learningToMeasure | March 20, 2010 | Reply

  3. [...] I say that any measure on an algebra is both monotone and subtractive. Indeed, since is an algebra, then is guaranteed [...]

    Pingback by Monotonicity of Measures « The Unapologetic Mathematician | March 22, 2010 | Reply

  4. [...] I say that a measure is continuous from above and [...]

    Pingback by Continuity of Measures « The Unapologetic Mathematician | March 23, 2010 | Reply

  5. [...] as Metric It turns out that a measure turns its domain into a sort of metric space, measuring the “distance” between two [...]

    Pingback by Measure as Metric « The Unapologetic Mathematician | March 24, 2010 | Reply

  6. [...] We’re going to want a modification of the notion of a measure. But before we introduce it, we have (of course) a few [...]

    Pingback by Outer Measures « The Unapologetic Mathematician | March 25, 2010 | Reply

  7. [...] a Measure to an Outer Measure Let be a measure in a ring (not necessarily an algebra) , and let be the hereditary -ring generated by . For every [...]

    Pingback by Extending a Measure to an Outer Measure « The Unapologetic Mathematician | March 26, 2010 | Reply

  8. [...] measure on a hereditary -ring 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 [...]

    Pingback by Sets Measurable by an Outer Measure I « The Unapologetic Mathematician | March 29, 2010 | Reply

  9. [...] additive. That is, if we define for , then is actually a measure. Even better, it’s a complete [...]

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

  10. [...] and Induced Measures If we start with a measure on a ring , we can extend it to an outer measure on the hereditary -ring . And then we can [...]

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

  11. [...] of a Measure At last we can show that the set function we defined on semiclosed intervals is a measure. It’s clearly real-valued and non-negative. We already showed that it’s monotonic, and [...]

    Pingback by An Example of a Measure « The Unapologetic Mathematician | April 16, 2010 | Reply

  12. [...] We’ve spent a fair amount of time discussing rings and -rings of sets, and measures as functions on such collections. Now we start considering how these sorts of constructions relate [...]

    Pingback by Measurable Spaces, Measure Spaces, and Measurable Functions « The Unapologetic Mathematician | April 26, 2010 | Reply

  13. [...] we don’t particularly care if the set where is false is itself measurable, although if is complete then all -negligible sets will be measurable. This sort of language is so common in measure theory [...]

    Pingback by Almost Everywhere « The Unapologetic Mathematician | May 13, 2010 | Reply

  14. [...] If is a.e. non-negative, then will also be non-negative, and so the indefinite integral is a measure. Since is integrable we see [...]

    Pingback by Extending the Integral « The Unapologetic Mathematician | June 21, 2010 | Reply

  15. [...] We continue what we started yesterday by extending the notion of a measure. We want something that captures the indefinite integrals of every function for which it’s [...]

    Pingback by Signed Measures « The Unapologetic Mathematician | June 22, 2010 | Reply

  16. [...] I say that each of these set functions — , , and — is a measure, and that . If is (totally) finite or -finite, then so are and , and at least one of them will [...]

    Pingback by Jordan Decompositions « The Unapologetic Mathematician | June 25, 2010 | Reply

  17. [...] real-valued functions on them. Given a function on a Boolean ring , we say that is additive, or a measure, -finite (on -rings), and so on analogously to the same concepts for set functions. We also say [...]

    Pingback by Functions on Boolean Rings and Measure Rings « The Unapologetic Mathematician | August 5, 2010 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 366 other followers

%d bloggers like this: