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

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

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 $\Sigma$.