How to prove that every set in a sigma ring is a subset of a countable union of sets in it generator? One of the "Remarks" in our lecture notes says that if $\mathcal G$ is non-empty family of sets and $\sigma_{\mathcal R}(\mathcal G)$ is the sigma ring generated by $\mathcal G$, then every set in $\sigma_{\mathcal R}(\mathcal G)$ is a subset of countable union of sets in $\mathcal G$. Why is this remark true.

Let
$$
\mathfrak X:=\{A: A \text{ is contained in a countable union of sets in } \mathcal G\}
$$
Observe that

*

*$\emptyset\in\mathfrak{X}$

*$\mathfrak{X}$ is closed under countable unions.

*$\mathfrak X$ is closed under differences (in fact, if $B\subset A\in \mathfrak X$, then $B\in\mathfrak X$).

*$\mathcal G\subset \mathfrak{X}$
Do you see how to complete the argument?
