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.

 

[blue_raver22]

I cannot provide a definitive answer as this question is specific to measure theory and may require expertise in that area. However, I can provide some insights and suggestions for further clarification.

Based on the definitions provided, it seems that the concept of an open cover and the concept of a σ-finite measure are related but not equivalent. An open cover is a collection of sets that covers a given set, while a σ-finite measure is a measure that can be decomposed into a countable union of finite sets. In other words, an open cover is a property of a set, while a σ-finite measure is a property of a measure.

It is possible that for a given set X, there exists an open cover \{ E_\alpha\}_{\alpha \in I} such that X \subset \bigcup_{\alpha \in I} E_\alpha and each E_\alpha has a finite measure. However, this does not necessarily mean that X is σ-finite. This is because a set can have an uncountable number of open covers, and it is not necessary that all of these covers will have the property of σ-finite measure.

To answer your question, it is not accurate to say that μ is σ-finite if there exists an open cover for X such that μ(E_i)<∞ for each i. This statement may hold true in some cases, but it is not a general definition for σ-finite measures.

My suggestion would be to consult with a mathematician or a textbook on measure theory for a more precise understanding of these concepts. Also, it may be helpful to look at some examples and counterexamples to gain a better understanding of the relationship between open covers and σ-finite measures.