# properties for measure

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

1. 1.

If $A,B\in\mathcal{B}$, and $A\subset B$, then $\mu(A)\leq\mu(B)$.

2. 2.

If $A,B$ in $\mathcal{B}$, $A\subset B$, and $\mu(A)<\infty$, then

 $\mu(B\setminus A)=\mu(B)-\mu(A).$
3. 3.

For any $A,B$ in $\mathcal{B}$, we have

 $\mu(A\cup B)+\mu(A\cap B)=\mu(A)+\mu(B).$
4. 4.

Subadditivity: If $\{A_{i}\}_{i=1}^{\infty}$ is a collection of sets from $\mathcal{B}$, then

 $\mu\big{(}\bigcup_{i=1}^{\infty}A_{i}\big{)}\leq\sum_{i=1}^{\infty}\mu(A_{i}).$
5. 5.

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

 $\mu\big{(}\bigcup_{i=1}^{\infty}A_{i}\big{)}=\lim_{i\to\infty}\mu(A_{i}).$
6. 6.

Continuity from above: If $\{A_{i}\}_{i=1}^{\infty}$ is a collection of sets from $\mathcal{B}$ such that $\mu(A_{1})<\infty$, and $A_{i}\supset A_{i+1}$ for all $i$, then

 $\mu\big{(}\bigcap_{i=1}^{\infty}A_{i}\big{)}=\lim_{i\to\infty}\mu(A_{i}).$

Remarks In (2), the assumption $\mu(A)<\infty$ assures that the right hand side is always well defined, i.e., not of the form $\infty-\infty$. 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

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

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

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

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

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

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 $\{D_{i}\}_{i=1}^{\infty}$ as

 $D_{1}=A_{1},\,\,\,\,\,\,\,\,D_{i}=A_{i}\setminus\bigcup_{k=1}^{i-1}A_{k}.$

Now $D_{i}\cap D_{j}=\emptyset$ for $i, so $\{D_{i}\}$ is a sequence of disjoint sets. Since $\cup_{i=1}^{\infty}D_{i}=\cup_{i=1}^{\infty}A_{i}$, and since $D_{i}\subset A_{i}$, we have

 $\displaystyle\mu(\bigcup_{i=1}^{\infty}A_{i})$ $\displaystyle=$ $\displaystyle\mu(\bigcup_{i=1}^{\infty}D_{i})$ $\displaystyle=$ $\displaystyle\sum_{i=1}^{\infty}\mu(D_{i})$ $\displaystyle\leq$ $\displaystyle\sum_{i=1}^{\infty}\mu(A_{i}),$

and property (4) follows.

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

## References

Title properties for measure PropertiesForMeasure 2013-03-22 13:45:28 2013-03-22 13:45:28 matte (1858) matte (1858) 8 matte (1858) Theorem msc 60A10 msc 28A10