The Unapologetic Mathematician

Mathematics for the interested outsider


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


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}.

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


  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

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