Assume my underlying space being
$E$ being separable, i.e. $\forall \epsilon > 0$ there exists a sequence in
$(a_i)_{i \in \mathbb{N}} \in E$ such that $E \subset \bigcup U_\epsilon(a_i)$, where $U_\epsilon $ is the $\epsilon$-neighbourhood of $a_i$.
Let $\mu$ be a $\sigma$-finite measure on $E$, i.e; there exists a countable sequence of sets $B_i$ covering $E$ and being of finite measure, $\mu(B_i) < \infty$.
Does this imply boundedly finitess, i.e. every bounded set $B$ has finite measure?
Does boundedly finitness of $\mu$ imply $\sigma$-finitness?
Here are my thoughts about this.
Boundedly finitness $\Rightarrow \sigma$-finiteness:
As $E$ can be covered by the bounded $U_\epsilon(a_i)$ we get $\sigma$-finitness easily.
$\sigma$-finiteness $\Rightarrow$ Boundedly finitness:
Let $B \subset E$ be bounded. $B$ is covered by a finite number of the $B_i$, which all have finite measure. Hence, $\mu(B) < \infty$.
To me it is not clear, why $B$ is covered by a finite number of the $B_i$. Is this really true or do I risk running into some pathological counterexample?
Best Answer
Your question seems to make some assumptions on $E$ - a measurable space is not a priori a metric space.
And no, a $\sigma$-finite measure on a metric space need not be boundedly finite - for instance because I can just change any metric to be bounded.
For example, you can equip $\mathbb{R}$ with the metric $d(x,y)=\min\{1,|x-y|\}$. It's not hard to check that this is a metric which is equivalent to the standard metric. However, under this metric, $\mathbb{R}$ is a bounded set which has infinite Lesbegue measure. Of course, the Lesbegue measure is $\sigma$-finite.