I have a statement of a Theorem (possibly some formulation of the Hahn-Kolmogarov theorem?) in my notes as follows:
Let $\mathfrak{A}$ be an algebra of subsets of $X$ and let $\mu_1, \mu_2$ be finite measures on $\sigma(\mathfrak{A})$. If $\mu_1(A) = \mu_2(A)$ for all $A \in \mathfrak{A}$, then $\mu_1 = \mu_2$.
The sketch of the proof I have is this:
Define $\mathfrak{M} = \{A \in \sigma(\mathfrak{A}) : \mu_1(A) = \mu_2(A)\}$. Then $\mathfrak{M}$ is a monotone collection containing $\mathfrak{A}$. Let $m(\mathfrak{A})$ denote the the monotone collection generated by $\mathfrak{A}$. Then $\mathfrak{M} \supseteq m(\mathfrak{A}) = \sigma(\mathfrak{A})$.
I’m following all of this.
However, then my notes say:
The result follows from the fact that $\mathfrak{M} = \sigma(\mathfrak{A})$.
I don’t see how that gives the result we need. I would appreciate any clarity on this!