Show a Carathéodory-constructed measurable set is almost in the underlying algebra

This is easy to do for finite-measure sets, but for the infinite case I can't seem to get a solution...

Let $\mathcal{A}$ be an algebra of subsets of $X$ (i.e. closed under FINITE unions and intersections, and complements, with $\emptyset,X\in\mathcal{A}$). Let $\mu$ be the Carathéodory-constructed measure generated by $\mathcal{A}$, and let $E$ be a $\mu$-measurable with $\mu(E)=\infty$. Suppose $\mathcal{A}$ is $\sigma$-finite by this measure, i.e. that there is a countable sequence $A_n\in\mathcal{A}$ with $\mu(A_n)<\infty)$ and $\bigcup A_n=X$. I must show that given $\epsilon>0$ there is $A\in\mathcal{A}$ such that $\mu(E\Delta A)<\epsilon$. [Recall that $E\Delta A=(A\cap E^c)\cup(A^c\cap E)$.]

As I said, the result holds easily when $\mu(E)<\infty$. But for $\mu(E)=\infty$, I'm stuck. Any help would be much appreciated. Thanks!