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

1. Monotonicity: If $A,B\in \cB$ , and $A\subset B$ , then $\mu(A)\le \mu(B)$ .
2. If $A,B$ in $\cB$ , $A\subset B$ , and $\mu(A)< \infty$ , then $$\mu(B\setminus A) = \mu(B)-\mu(A).$$
3. For any $A,B$ in $\cB$ , we have $$ \mu(A\cup B)+\mu(A\cap B) = \mu(A) + \mu(B).$$
4. Subadditivity: If $\{A_i\}_{i=1}^\infty$ is a collection of sets from $\cB$ , then $$\mu\big(\bigcup_{i=1}^\infty A_i\big) \le \sum_{i=1}^\infty \mu(A_i).$$
5. Continuity from below: If $\{A_i\}_{i=1}^\infty$ is a collection of sets from $\cB$ 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. Continuity from above: If $\{A_i\}_{i=1}^\infty$ is a collection of sets from $\cB$ 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 \begin{eqnarray*} \mu(B) = \mu(A\cup (B\setminus A)) = \mu(A) + \mu(B\setminus A). \end{eqnarray*}Since $\mu(B\setminus A)\ge 0$ , 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 \begin{eqnarray*} \mu(A\cup B) &=& \mu(A)+\mu(B\setminus A)\\ &\le & \mu(A)+\mu(B). \end{eqnarray*}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 \begin{eqnarray*} \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), \end{eqnarray*}where, in the second equality we have used properties for the symmetric set difference, 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 $i<j$ , 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 \begin{eqnarray*} \mu(\bigcup_{i=1}^\infty A_i) &=& \mu(\bigcup_{i=1}^\infty D_i) \\ &=& \sum_{i=1}^\infty \mu(D_i) \\ &\le& \sum_{i=1}^\infty \mu(A_i), \end{eqnarray*}and property (4) follows.

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

Bibliography

1

G.B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd ed, John Wiley & Sons, Inc., 1999.

2

A. Mukherjea, K. Pothoven, Real and Functional analysis, Plenum press, 1978.

3

D.L. Cohn, Measure Theory, Birkhäuser, 1980.

4

A. Friedman, Foundations of Modern Analysis, Dover publications, 1982.