Exercise 1.4.22 (Countable combinations of measures)

For each measure we verify the properties from the Definition 1.4.27.

(i)

We have $\mathit{c\mu }(\varnothing )=c\cdot \mu (\varnothing )=c\cdot 0=0$. Whenever ${\left(E\right)}_{n\in \mathbb{N}}\in B$ are countable sequence of disjoint measurable sets, then we have $$(\mathit{c\mu })\left(\underset{n=1}{\overset{\infty }{\bigcup }}{E}_n\right)=c\cdot \mu \left(\underset{n=1}{\overset{\infty }{\bigcup }}{E}_n\right)=c\cdot \underset{n=1}{\overset{\infty }{\sum }}\mu (E_n)=\underset{n=1}{\overset{\infty }{\sum }}(\mathit{c\mu })(E_n).$$

(ii)

Similarly, we have $\sum <!--\; FUNCTION\; APPLICATION\; -->\mu (\varnothing )={\sum }_{n=1}^{\infty }{\mu }_n(\varnothing )={\sum }_{n=1}^{\infty }0=0$. Furthermore, whenever $E_1,E_2,\dots \in B$ are countable sequence of disjoint measurable sets, by Tonelli-Fubini theorem (see p. xiv) we have $$\sum \mu \left(\underset{n=1}{\overset{\infty }{\bigcup }}{E}_n\right)=\underset{k=1}{\overset{\infty }{\sum }}{\mu }_k\left(\underset{n=1}{\overset{\infty }{\bigcup }}{E}_n\right)=\underset{k=1}{\overset{\infty }{\sum }}\underset{n=1}{\overset{\infty }{\sum }}{\mu }_k(E_n)=\underset{n=1}{\overset{\infty }{\sum }}\underset{k=1}{\overset{\infty }{\sum }}{\mu }_k(E_n)=\underset{n=1}{\overset{\infty }{\sum }}\sum \mu (E_n).$$