properties for measure

Theorem [1, 2, 3, 4] Let $(E,ℬ,\mu )$ be a measure space, i.e., let $E$ be a set, let $ℬ$ be a $\sigma$-algebra of sets in $E$, and let $\mu$ be a measure on $ℬ$. Then the following properties hold:

1. Monotonicity: If $A,B\in ℬ$, and $A\subset B$, then $\mu (A)\le \mu (B)$.

2. If $A,B$ in $ℬ$, $A\subset B$, and , then

   $$\mu (B\setminus A)=\mu (B)-\mu (A).$$

3. For any $A,B$ in $ℬ$, we have

   $$\mu (A\cup B)+\mu (A\cap B)=\mu (A)+\mu (B).$$

4. Subadditivity: If ${\{A_i\}}_{i=1}^{\mathrm{\infty }}$ is a collection of sets from $ℬ$, then

   $$\mu \left(\bigcup _{i=1}^{\mathrm{\infty }}{A}_i\right)\le \sum _{i=1}^{\mathrm{\infty }}\mu (A_i).$$

5. Continuity from below: If ${\{A_i\}}_{i=1}^{\mathrm{\infty }}$ is a collection of sets from $ℬ$ such that $A_i\subset A_{i+1}$ for all $i$, then

   $$\mu \left(\bigcup _{i=1}^{\mathrm{\infty }}{A}_i\right)=\underset{i\to \mathrm{\infty }}{lim}\mu (A_i).$$

6. Continuity from above: If ${\{A_i\}}_{i=1}^{\mathrm{\infty }}$ is a collection of sets from $ℬ$ such that , and $A_i\supset A_{i+1}$ for all $i$, then

   $$\mu \left(\bigcap _{i=1}^{\mathrm{\infty }}{A}_i\right)=\underset{i\to \mathrm{\infty }}{lim}\mu (A_i).$$

Remarks In (2), the assumption assures that the right hand side is always well defined, i.e., not of the form $\mathrm{\infty }-\mathrm{\infty }$. Without the assumption we can prove that $\mu (B)=\mu (A)+\mu (B\setminus A)$ (see below). In (3), it is tempting to move the term $\mu (A\cap B)$ to the other side for aesthetic reasons. However, this is only possible if the term is finite.

Proof. For (1), suppose $A\subset B$. We can then write $B$ as the disjoint union $B=A\cup (B\setminus A)$, whence

$\mu (B)=\mu (A\cup (B\setminus A))=\mu (A)+\mu (B\setminus A).$

Since $\mu (B\setminus A)\ge 0$, the claim follows. Property (2) follows from the above equation; since , we can subtract this quantity from both sides. For property (3), we can write $A\cup B=A\cup (B\setminus A)$, whence

$\mu (A\cup B)$ $=$ $\mu (A)+\mu (B\setminus A)$

$\le$ $\mu (A)+\mu (B).$

If $\mu (A\cup B)$ is infinite, the last inequality must be equality, and either of $\mu (A)$ or $\mu (B)$ must be infinite. Together with (1), we obtain that if any of the quantities $\mu (A),\mu (B),\mu (A\cap B)$ or $\mu (A\cup B)$ is infinite, both sides in the equation are infinite and the claim holds. We can therefore without loss of generality assume that all quantities are finite. From $A\cup B=B\cup (A\setminus B)$, we have

$$\mu (A\cup B)=\mu (B)+\mu (A\setminus B)$$

and thus

$$2\mu (A\cup B)=\mu (A)+\mu (B)+\mu (A\setminus B)+\mu (B\setminus A).$$

For the last two terms we have

$\mu (A\setminus B)+\mu (B\setminus A)$ $=$ $\mu ((A\setminus B)\cup (B\setminus A))$

$=$ $\mu ((A\cup B)\setminus (A\cap B))$

$=$ $\mu (A\cup B)-\mu (A\cap B),$

where, in the second equality we have used properties for the symmetric set difference (http://planetmath.org/SymmetricDifference), and the last equality follows from property (2). This completes the proof of property (3). For property (4), let us define the sequence ${\{D_i\}}_{i=1}^{\mathrm{\infty }}$ as

$$D_1=A_1,D_i=A_i\setminus \bigcup _{k=1}^{i-1}{A}_k.$$

Now $D_i\cap D_j=\mathrm{\varnothing }$ for , so $\{D_i\}$ is a sequence of disjoint sets. Since ${\cup }_{i=1}^{\mathrm{\infty }}{D}_i={\cup }_{i=1}^{\mathrm{\infty }}{A}_i$, and since $D_i\subset A_i$, we have

$\mu ({\displaystyle \bigcup _{i=1}^{\mathrm{\infty }}}{A}_i)$ $=$ $\mu ({\displaystyle \bigcup _{i=1}^{\mathrm{\infty }}}{D}_i)$

$=$ ${\displaystyle \sum _{i=1}^{\mathrm{\infty }}}\mu (D_i)$

$\le$ ${\displaystyle \sum _{i=1}^{\mathrm{\infty }}}\mu (A_i),$

and property (4) follows.

TODO: proofs for (5)-(6).

Title	properties for measure
Canonical name	PropertiesForMeasure
Date of creation	2013-03-22 13:45:28
Last modified on	2013-03-22 13:45:28
Owner	matte (1858)
Last modified by	matte (1858)
Numerical id	8
Author	matte (1858)
Entry type	Theorem
Classification	msc 60A10
Classification	msc 28A10