Exercise 1.6

Proof. We need to verify the conditions in Definition $1.3(a)$. $(i)$ $X\in 𝔐$ since $X^c=\varnothing$ is at most countable. $(\mathit{ii})$ If $E\in 𝔐$, then $E^c\in 𝔐$ since one of $E,E^c$ is at most countable. $(\mathit{iii})$ Suppose $E_n\in 𝔐$. If all $E_n$ are at most countable, then $\bigcup <!--\; FUNCTION\; APPLICATION\; -->E_n$ is at most countable, hence $\bigcup <!--\; FUNCTION\; APPLICATION\; -->E_n\in 𝔐$. Otherwise, there is some $E_{n_0}$ such that $E_{n_0}$ is not at most countable but $E_{n_0}^c$ is. Then, we have

since this intersection is contained in $E_{n_0}^c$.

We now want to show $\mu$ is a measure. We need to verify countable additivity. Suppose $\{E_n\}$ is a disjoint countable collection of members of $𝔐$. If all the $E_n$ are countable, then we are done since the countable union of countable sets is countable, hence

So suppose some $E_{n_0}$ is not countable, but $E_{n_0}^c$ is. Then, by the above ${\left(\bigcup <!--\; FUNCTION\; APPLICATION\; -->E_n\right)}^c$ is countable, so $\mu \left(\bigcup <!--\; FUNCTION\; APPLICATION\; -->E_n\right)=1$. Moreover, every other $E_n$ is contained in $E_{n_0}^c$, which is countable, hence every other $E_n$ is countable, and so $\sum <!--\; FUNCTION\; APPLICATION\; -->\mu (E_n)=\mu (E_{n_0})=1$.

We now describe the corresponding measurable functions and their integrals. Let $f:X\to R^1$ be measurable. Then, $f^{-1}([n,n+1])$ is measurable for each $n$, and since $X$ is uncountable, $f^{-1}([n,n+1])$ must have countable complement for some $n$. Similarly, we can keep partitioning $[n,n+1]$ into smaller and smaller intervals of length $1∕2^k$, and the preimage of one of these sets must have countable complement. Thus, taking the intersection of all these sets, $f^{-1}(x)$ has countable complement for some $x\in R^1$. In this way, we see $f$ is a constant $y\in R^1$ for all but countably many points in $X$. Finally, ${\int }_{X}f\mathit{d\mu }=y$. □