# Show that the difference of two Borel sets is itself a Borel set

I am currently in a Measure Theory class and we are going over Borel sets. I am having difficulty with the following proof: show that for any two Borel sets $A,B$, the difference $A-B$ is a Borel set. I feel like this is pretty basic and I am just missing something obvious. I understand that the collection of Borel sets contains the null set, all $n$-dimensional intervals, all open sets $G_ \delta$ and all closed sets $F_ \sigma$, as well as all countable intersections of open sets and countable unions of closed sets. I am fairly certain the direction that I should head is in noting that if $A$ is open, $A^c$ must be closed, and that $A-B=A \cap B^c$, but other than that, I am kind of lost.

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What is the definition of the Borel sets you use? With most standard definitions, the result is trivial. – Michael Greinecker Oct 17 '12 at 9:55
Yes that is why I am having difficulty. The way Borel set is defined in the class is "the sigma-algebra generated by the collection in R^n of all open sets". Thank you for taking a look @Michael – Angelo Christophell Oct 18 '12 at 5:56

This holds for every $\sigma$-algebra. Let $A$ and $B$ be elements of a $\sigma$-algebra. We have $A-B=A\cap B^C= (A^C\bigcup B)^C$. Since every $\sigma$-algebra is closed under complements and countable unions, this implies that $A-B$ is in the $\sigma$-algebra.
It gets even easier when you use the fact that a $\sigma$-algebra is closed under countable intersections, which can easily be proven using the De Morgan's laws in their generalized form.