Counting Measure not $\sum$-finite

Question

Exercise 3.13e in Cinlar's Probability and Stochastics:

Consider the measurable space $(E,B(E))$, where $E=[0,1]$ and $B(E)$ is the set of all Borel subsets of $E$.

Show that the counting measure $\mu$ on it is not $\sigma$-finite and also not $\sum$-finite, where $\sum$-finite means that there is a sequence of finite measures $\mu_1, \mu_2, ...$ such that $\mu=\sum{\mu_i}$.

I proved the not $\sigma$-finite part by showing that that implies the countability of $[0,1]$, a contradiction, but I'm not sure how to prove non-$\sum$-finiteness.

Answer 1

I think i may have a counter example. I am not sure if I am right, but I will post it here:

Define

$\forall A\in \mathcal{B}([0,1])$, $\delta_x(A)=\begin{cases} 1 & \text{if $x\in A$}\\ 0 & \text{if $x\notin A$}\\ \end{cases}$

Let $(\delta_x)_{x\in[0,1]}$ be a sequence of functions. Note that $\forall x\in[0,1]$, $\delta_x$ is a finite measure.

In the question, $\mu$ is the counting measure with $D=[0,1]$. By definition of counting measure with $D=[0,1]$, $\mu=\sum\limits_{x\in[0,1]}\delta_x$. Thus $\mu$ is $\Sigma$-finite by definition of $\Sigma$-finite.

What is the problem with my example?

Answer 2

Suppose for contradiction that $\mu = \sum_i \mu_i$, where the $\mu_i$ are finite. Then we can do the following: For each $x \in [0, 1]$, we know that $\sum_i \mu_i (\{x\} ) = 1$. Let $$F_i = \{ x : \mu_i(\{x\}) > 0\}.$$ We know that least one of these is uncountable, since $\bigcup_i F_i = [0, 1]$; after all, if $x \not \in \bigcup_i F_i$, then $\mu_i (\{x\}) = 0$ for all $i$, meaning $\mu (\{x\}) = 1 = \sum_i \mu_i (\{x\}) = 0$. But $[0, 1]$ is uncountable, and the union is indexed by a countable set, so some $F_i$ must be uncountable. Call this $i_0$.

Now let $E_k = \{ x \in F_{i_0}: \mu_{i_0} (\{ x \}) > 1/k \}$. Since $F_{i_0} = \bigcup_k E_k$ is uncountable, we know that some $E_{k_0}$ is uncountable.

Now, there’s no guarantee that $E_{k_0}$ is Borel, but it’ll contain a countable infinite subset, which is Borel. Let $(x_n)_{n = 1}^\infty$ be a sequence of distinct elements in $E_{k_0}$. Then $\{x_n : n \in \mathbb{N} \}$ will be a Borel set, and $\mu_{i_0} (\{ x_n : n \in \mathbb{N} \} ) \geq \sum_{n = 1}^\infty 1/k_0 = \infty$, a contradiction.

Answer 3

Suppose that $\mu$ is $\sigma$-finite, $\mu = \sum_n \mu_n$ with each $\mu_n$ finite. Now, for fixed $n$ and any natural $m$, we know that

$$ \left\{x \in [0, 1] : \mu_n(\{x\}) > \frac{1}{m}\right\} $$ also has at most $m \mu_n([0, 1])$ elements, i.e. is finite. And this implies that $$ \left\{x \in [0, 1] : \mu_n(\{x\}) > 0\right\} = \bigcup_m\left\{x \in [0, 1] : \mu_n(\{x\}) > \frac{1}{m}\right\} $$ is at most countable, which in turn implies that $$ A = \bigcup_n \left\{x \in [0, 1] : \mu_n(\{x\}) > 0\right\} $$ is at most countable. Since $[0, 1]$ is uncountable, then, we can pick $x$ to be in $[0, 1]\backslash A$, and then find a contradiction: by definition of $A$, we must have $\sum_n \mu_n(\{x\}) = 0$, but $\mu(\{x\}) = 1$.

(Based on this real analysis answer and the hint to Exercise I.3.14 in Çinlar's Probability and Stochastics.)