Question on Definition of Cover of a Set

[BrainHurts]

So when we have an open cover of a set X means we have a collection of sets $\{ E_\alpha\}_{\alpha \in I}$

such that $X \subset \bigcup_{\alpha \in I} E_\alpha$.

My question comes from measure theory, on the question of finite $\sigma$ -measures,

The definition I'm readying says $\mu$ is $\sigma$ - finite if there exists sets $E_i \in \mathcal{A}$ for $i = 1,2, ...$ such that $\mu(E_i) < \infty$ for each $i$ and $X = \bigcup_{i=1}^\infty E_i$.

Can I say that μ is σ - finite if there exists an open cover for X such that μ(E_i)< ∞ for each i? Is my understanding correct?

 

[gopher_p]

Make it a countable open cover, and you're good to go.

 

[jgens]

It merits note that the converse claim "If μ is σ-finite, then there exists a countable open cover {E_k} with μ(E_k) < ∞" (where X is a topological space and μ is a measure over the Borel algebra) is generally false.

 

[economicsnerd]

It merits note that the converse claim "If μ is σ-finite, then there exists a countable open cover {E_k} with μ(E_k) < ∞" (where X is a topological space and μ is a measure over the Borel algebra) is generally false.

As an example, consider the "rational counting" measure $\mu:\mathcal B_{\mathbb R}\to [0,\infty]$ on the real line. So $\mu(A) = |A\cap\mathbb Q|$. The measure $\mu$ is $\sigma$-finite, as witnessed by the countable Borel cover $\{\{r\}:\enspace r\in\mathbb Q\}\cup\{\mathbb R\setminus\mathbb Q\}$. However, every nonempty open set has infinite $\mu$-measure.