Suppose that $(X,\mathcal{A})$ is a measurable space and let $\mathcal{A}^0$ be an algebra generating $\mathcal{A}$. For each $\epsilon>0$ and $A\in \mathcal{A}$, one can find $A^0\in\mathcal{A}^0$ such that $\mu(A\Delta A^0)<\epsilon$, where $\mu$ is a finite measure on $(X,\mathcal{A})$. Now, if we're given another finite measure $\nu$ on $(X,\mathcal{A})$, can one find $A^0\in\mathcal{A}^0$ such that both $\mu(A\Delta A^0)<\epsilon_1$ and $\nu(A\Delta A^0)<\epsilon_2$ for some $\epsilon_1,\epsilon_2>0$? (A similar argument is used in the accepted answer here.)
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Assuming the measures are non-negative:
Consider the non-negative measure $\eta$ given by $\eta = \mu + \nu$. Do you see how you can proceed from here?
In full detail, given $A\in \mathcal{A}$ we can find $A^{0} \in \mathcal{A}^{0}$ such that this new measure is small on $A\Delta A^{0}$: for $\varepsilon = \min\{\varepsilon_1, \varepsilon_2\}$ we have that: $$\varepsilon > \eta(A\Delta A^{0}) =\mu(A\Delta A^{0}) + \nu(A\Delta A^{0}).$$ Now, as both measures are non-negative the previous inequality implies what you wanted to prove.
Nicolas Agote
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