1

I've seen people using arguments along the lines of the following one, but never a proof or a detailed statement. Can anyone help me find one?

Let $\mathcal{E}$ be a vector space of bounded functions $X\to\Bbb{R}$ containing $1$, and suppose that it is closed under products and monotone limits. Let $\Sigma$ be the collection of all subsets $E$ of $X$ such that the indicator function $1_E$ is in $\mathcal{E}$. Then $\Sigma$ is a $\sigma$-algebra, and every function in $\mathcal{E}$ is $\Sigma$-measurable.

First of all, is the statement correct to begin with?

I can see why we have a $\sigma$-algebra. For the rest, I feel like the Weierstrass approximation theorem should help, but I'm stuck.

(A reference would be welcome too, especially if this statement, or a closely related one, has a name.)

Note: this is not the usual monotone class theorem/pi-lambda theorem. We are asking whether every function in $\mathcal{E}$ is $\Sigma$-measurable, not if every $\Sigma$-measurable is in $\mathcal{E}$. (That's why we require that $\mathcal{E}$ is closed under products.)

geodude
  • 8,357
  • I think we cannot use Weierstrass. The functions are not continuous, and indeed $X$ is not a topological space. – GEdgar May 21 '23 at 17:25
  • If $f \in \mathcal E$, then $f$ is $\Sigma$-measurable. Hint. Start with a real number $t$. We want to reach the indicator of the set ${x : f(x) > t}$ starting with $f$, with $1$, and using vector operations, products, monotone limits. – GEdgar May 21 '23 at 17:29
  • See here for a similar statement, with reference for a proof: https://math.stackexchange.com/a/47521/151552 – PhoemueX May 22 '23 at 18:34
  • @PhoemueX It is not quite that. See the edit. – geodude May 23 '23 at 08:24

0 Answers0