If $\mu_1\le\mu_2$ on an algebra, then $\mu_1\le\mu_2$ on the $\sigma$-algebra generated by that algebra

Question

We suppose that $\mu_1$ and $\mu_2$ are finite measures on a $\sigma$-algebra $\Sigma$ generated by the algebra $\Sigma_0$.

My question is how to prove that if $\mu_1(A) \leq \mu_2(A)$ holds for all $A$ in $\Sigma_0$, then it holds for all $A$ in $\Sigma$.

Answer 1

We can use an approximation argument: if $A\in \Sigma$ and $\varepsilon>0$, take $A'\in\Sigma_0$ such that $(\mu_1+\mu_2)(A\Delta A')\lt\varepsilon$.

Then $$\mu_1(A)=\mu_1(A')+(\mu_1(A)-\mu_1(A'))\leqslant \mu_2(A')+(\mu_1(A)-\mu_1(A'))\lt \mu_2(A')+\varepsilon\\ \leqslant \mu_2(A)+2\varepsilon.$$ We conclude, as $\varepsilon$ was arbitrary.