# The Unapologetic Mathematician

## Monotonicity of Measures

First, we make a couple of definitions. A set function $\mu$ on a collection of sets $\mathcal{E}$ is “monotone” if whenever we have $E_1$ and $E_2$ in $\mathcal{E}$ with $E_1\subseteq E_2$, then $\mu(E_1)\leq\mu(E_2)$. We say that $\mu$ is “subtractive” if whenever further $E_2\setminus E_1\in\mathcal{E}$ and $\lvert\mu(E_1)\rvert<\infty$, then $\mu(E_2\setminus E_1)=\mu(E_2)-\mu(E_1)$.

Now I say that any measure on an algebra $\mathcal{A}$ is both monotone and subtractive. Indeed, since $\mathcal{A}$ is an algebra, then $E_2\setminus E_1$ is guaranteed to be in $\mathcal{A}$ as well, and it’s disjoint from $E_1$. And so we see that

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

Since $\mu$ is non-negative, we must have $\mu(E_2)\geq\mu(E_1)$, and so $\mu$ is monotone. And if $\mu(E_1)$ is finite, then we can subtract it from both sides of this equation to show that $\mu$ is subtractive as well.

Next, say $\mu$ is a measure on an algebra $\mathcal{A}$. If we have a set $A\in\mathcal{A}$ and a finite or (countably) infinite sequence of sets $\{A_i\}$ so that $A\subseteq\bigcup_iA_i$, then we have the inequality $\mu(A)\leq\sum_i\mu(A_i)$. To show this, we’ll invoke a very useful trick: if $\{F_i\}$ is a sequence of sets in an algebra $A$, then there is a pairwise disjoint sequence $\{G_i\}$ of sets so that each $G_i\subseteq F_i$ and $\bigcup_iG_i=\bigcup_iF_i$. Indeed, we define

$\displaystyle G_i=F_i\setminus\bigcup\limits_{1\leq j

That is, $G_1$ is the same as $F_1$, and after that point each $G_i$ is everything in $F_i$ that hasn’t been covered already.

So we start with $B_i=A\cap A_i$, and come up with a new sequence $\{C_i\}$. The (disjoint) union of the $C_i$ is, $A$ like the union of the $B_i$. Then, by the countable additivity of $\mu$, we have $\mu(A)=\sum_i\mu(C_i)$. But the monotonicity says that $\mu(C_i)\leq\mu(B_i)$, and also that $\mu(B_i)\leq\mu(A_i)$, and so $\mu(A)\leq\sum_i\mu(A_i)$.

On the other hand, if $\{A_i\}$ is a pairwise disjoint and we have $\bigcup_iA_i\subseteq A$, then $\sum_i\mu(A_i)\leq\mu(A)$. Indeed, if $\{A_i\}$ is a finite sequence, then the union $\bigcup_iA_i$ is in $\mathcal{A}$. Monotonicity and finite additivity then tell us that

$\displaystyle\sum\limits_i\mu(A_i)=\mu\left(\bigcup\limits_iA_i\right)\leq\mu(A)$

However if $\{A_i\}$ is an infinite sequence, then what we just said applies to any finite subsequence. Then the sum $\sum_{i=1}^\infty\mu(A_i)$ is the limit of the sums $\sum_{i=1}^n\mu(A_i)$. Each of these is less than or equal to $\mu(A)$, and so their limit must be as well.

March 22, 2010 Posted by | Analysis, Measure Theory | 8 Comments