The Unapologetic Mathematician

Mathematics for the interested outsider

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<i}F_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.

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March 22, 2010 - Posted by | Analysis, Measure Theory

8 Comments »

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