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.)