1

Consider the following theorem for equivalence of measures:

Let $(E, \mathcal{A})$ be a measureable space with $\mathcal{A} = \sigma(\mathcal{G})$ for a ($\cap$-stable family $\mathcal{G}$) and $\mu, \nu$ be measures on that spaces. If there exists a sequence $(G)_{n \in \mathbb{N}} \in \mathcal{G}$ which ascends to $E$ (ie. $G_n \uparrow E$) and $\mu(G_n), \nu(G_n) < \infty$ for all $n$ then $$\mu(G) = \nu(G) ~~\forall G \in \mathcal{G} ~\implies~ \mu = \nu ~\text{on}~ \mathcal{A}.$$

Here are my questions:

  1. Equivalently one can say, if ($\sigma$-finite) measures agree on $\mathcal{G}$, then they are already equal on $\mathcal{A}$. How can I translate this (more intuitive statement) to the "there exists a sequence" statement?

  2. Why is $\mu(G_n), \nu(G_n) < \infty$ required? Can someone give a pathological example if this requirement is not fullfiled?

Thank you.

elfeck
  • 427
  • Your "intuitive" statement is not equivalent to $\sigma$-finiteness. $\sigma$-finiteness is a required condition for the statement to hold, not equivalent to what follows. – user160738 Feb 24 '17 at 16:19
  • Okay, I see. I edited my post accordingly. The same questions remain, still – elfeck Feb 24 '17 at 16:31
  • 2
    "there exists a sequence" statement is literally what is meant by $\sigma$-finiteness, so I don't see what you mean by that – user160738 Feb 24 '17 at 16:32

1 Answers1

1

For (2), let $E = \mathbb{N}$ and $\mathcal{A}=2^{\mathbb{N}}$. Let $\mathcal{G} = \{\{n, n+1, n+2, \dots\} : n \ge 0\}$. Then $\mathcal{A} =\sigma(\mathcal{G})$ and $\mathcal{G}$ is stable under intersection. There is a sequence in $\mathcal{G}$ ascending to $E$; in particular, $\mathbb{N} \in \mathcal{G}$ so we can take $G_n = \mathbb{N}$. Let $\mu$ be counting measure and $\nu = 2\mu$. It is true that $\mu(G) = \nu(G) = \infty$ for all $G \in \mathcal{G}$, but it is not true that $\mu=\nu$.

Nate Eldredge
  • 97,710