# The Unapologetic Mathematician

## 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 $\mathcal{A}$.

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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