Why is the boolean closure of $F_{\sigma}$-sets not in $F_{\sigma}\cap G_{\delta}$?

Question

In the Borel hierarchy, why is the boolean closure of $F_{\sigma}$ or $G_{\delta}$ equal to $F_{\sigma \delta} \cap G_{\delta \sigma}$? If I take the complement of an element in $F_{\sigma}$ I got an element of $G_{\delta}$, and finite unions and intersection of elements in $F_{\sigma}$ are still in $F_{\sigma}$, so as I see it the boolean closure should be contained in $F_{\sigma} \cap G_{\delta}$?

Answer 1

For example, the union of an $F_\sigma$ and a $G_\delta$ might not be either $F_\sigma$ or $G_\delta$.

Answer 2

Just for the record, Boolean combinations of $F_\sigma$ sets do not give all the sets in $F_{\sigma\delta}\cap G_{\delta\sigma}$. Inside of $F_{\sigma\delta}\cap G_{\delta\sigma}$ sit $\omega_1$ classes of the difference hierarchy and the Boolean combinations of $F_\sigma$ sets are equal to the first $\omega$ levels of this hierarchy. Details are in Chapter 22.E of the book of A.S. Kechris "Classical Descriptive Set Theory".