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

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