Exercise 1.7.4 (Pre-measure)

Definition 1.7.7 is as follows:

We now verify the theorem hypotheses.

(i)

Suppose that ${\mu }_0(\varnothing )=0$. Then, for any $E_1,E_2\in B_0$ and $E_n=\varnothing$ for $n>2$ we can recover finite additivity by $${\mu }_0(E\cup F)={\mu }_0\left(\bigcup _{n=1}^{\infty }{E}_n\right)=\sum _{i=1}^{\infty }{\mu }_0(E_n)={\mu }_0(E_1)+{\mu }_0(E_2)+\sum _{i=3}^{\infty }{\mu }_0(E_n)={\mu }_0(E_1)+{\mu }_0(E_2).$$

(ii)

Notice that finite additivity of a pre-measure automatically automatically implies monotonicity as well: $${\mu }_0(E\cup F)={\mu }_0(E\cup (F∕E))={\mu }_0(E)+{\mu }_0(F∕E)\le {\mu }_0(E)+{\mu }_0(F).$$

By finite additivity and monotonicity, for each $N\in N$ we have

$${\mu }_0\left(\bigcup _{n=1}^{\infty }{E}_n\right)\ge {\mu }_0\left(\bigcup _{n=1}^N{E}_n\right)=\sum _{n=1}^N{\mu }_0\left(E_n\right).$$

Taking limits as $N\to \infty$ preserves the inequality

$${\mu }_0\left(\bigcup _{n=1}^{\infty }{E}_n\right)\ge \sum _{n=1}^{\infty }{\mu }_0\left(E_n\right).$$

Combined with the newly assumed countable subadditivity, this gives the desired result.

(iii)

We are looking for a function ${\mu }_0:B_0\to [0,+\infty ]$ such that

(i)

${\mu }_0(\varnothing )=0$,

(ii)

${\mu }_0\left({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{E}_n\right)\le {\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{\mu }_0(E_n)$ for any sequence of disjoint sets ${(F_n)}_{n\in N}$ in $B_0$, yet

(iii)

${\mu }_0\left({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{F}_n\right)\ne {\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{\mu }_0(F_n)$ for some sequence of disjoint sets ${(F_n)}_{n\in N}$ in $B_0$.

Consider the Jordan inner measure $m^{\ast ,(J)}(E)$ on a $\sigma$-algebra $L(R)$ of all Lebesgue measurable sets of $R$. Then

• $m^{\ast ,(J)}(E)$ is non-negative as it is a infimum over a set of non-negative elementary measures.

(i)

$m^{\ast ,(J)}(\varnothing )=0$, obviously.

(ii)

$m^{\ast ,(J)}(E)$ is subadditive by the properties of infimum

$$\inf <!--\; FUNCTION\; APPLICATION\; -->\left\{m(A):\begin{array}{c}\bigcup _{n=1}^{\infty }{E}_n\subseteq A\hfill \\ A\text{ is elementary}\hfill \end{array}\right\}\le \sum _{n=1}^{\infty }\inf <!--\; FUNCTION\; APPLICATION\; -->\left\{m(A):\begin{array}{c}{E}_n\subseteq A\hfill \\ A\text{ is elementary}\hfill \end{array}\right\}$$

(iii)

$m^{\ast ,(J)}(E)$ is not countably additive, as per Exercise 1.1.18 (4) for $q\in Q\cap [0,1]$ we have $$1=m^{\ast ,(J)}(Q\cap [0,1])\ne \sum _{q\in Q}{m}^{\ast ,(J)}(\{q\})=0.$$