Show that the set $E \subset \mathbb{R}$ being measurable is equivalent to E cleanly dividing any closed interval I (w.r.t the Lebesgue outer measure)

Question

This fact was stated without proof in a lecture, but I am struggling to justify why this is true.

If a set E is measurable, then, for any $A \subset \mathbb{R}$ we have $$m^*(A) = m^*(A \cap E) + m^*(A \cap E^{c})$$

Thus, if E is measurable, then since the above is true for any subset of the real line, E cleanly divides any closed interval.

I'm struggling to show the other way round though. I know that you can approximate any measurable set E by some closed set F $\subset$ E where $m^*(E \cap F^{c}) < \epsilon$ for any $\epsilon > 0$, but I don't think this helps, because I don't need the sets E to be closed, but the sets A in the above definiton.

Could anyone give some advice about a direction to work?

Thanks.