If a measure is a sum of finite measures then the underlying space must not necessarily be $\sigma$-finite

Question

Give a counterexample.

Let $(X,\mathcal{A})$ be a measurable space and $\mu$ a measure on $(X,\mathcal{A})$. If there are finite measures $(\mu_n)_{n \in \mathbb{N}}$ such that $\mu = \sum_{n \in \mathbb{N}}\mu_n$, then $\mu$ is $\sigma$-finite.

Let $(x_n)_{n \in \mathbb{N}}$ be a real sequence. Then $$\mu := \sum_{n \in \mathbb{N}} \delta_{x_n}$$ is a measure on $(\mathbb{R},\mathcal{P}(\mathbb{R}))$, where $\delta_x$ is the Dirca-measure in $x$. Furthermore, we have that $\mu$ is $\sigma$-finite if and only if there exists no $x \in \mathbb{R}$ such that $x_n = x$ for infinitely many $n \in \mathbb{N}$.

I wondered if there is any easier counterexample for the above statement since the one above is rather difficult to show. Any ideas?

Answer 1

$\sigma$-finiteness of a measure depends very much on the underlying $\sigma$-algebra. For instance if we consider the trivial $\sigma$-algebra $\mathcal{A} := \{\emptyset,\mathbb{R}\}$, then it is pretty obvious that $$\mu = \sum_{n \in \mathbb{N}} \delta_{x_n}$$ is not $\sigma$-finite (considered as a measure on $(\mathbb{R},\mathcal{A})$) for any choice of $(x_n)_{n \in \mathbb{N}}$.