Are these sets Borel-measurable? Are the following sets Borel-measurable and if so, what is the value of the measure? 
1) A = {(x,y) ∈ $[0,1]^2$| x and y rational}
2) B = {(x,y) ∈ $[0,1]^2$ | x or y rational}
3) C = {(x,y) ∈ $[0,1]^2$ | x and y irrational}
4) D = {(x,y) ∈ $[0,1]^2$ | x=y}
 A: Some hints to help you further in this.
1) Singletons are Borel-measurable and countable unions of Borel-measurable sets are Borel-measurable. Conclusion: countable sets (which are countable unions of singletons) are measurable. The Lebesgue measure of a singleton is $0$. What can you conclude then about the Lebesgue measure of a countable set?
2) What do you think: are sets of the form $\{x\}\times[0,1]$ or $\times[0,1]\times\{y\}$ Borel-measurable? If so then what is the Lebesgue measure on these sets? Note that $B$ can be written as a countable union of this sort of sets.
3) $C$ is the complement of $B$. What is your conclusion?
4) $D$ is closed. What can be concluded from that? Be aware that the Borel-measurable sets together form the $\sigma$-algebra generated by the open sets.

edit:
First let me emphasize what I said about Borel-measurable sets under 4). That implies directly that open (hence also their complements, the closed sets) are Borel-measurable.
Consequently singletons are measurable and for a countable set $A$ we have $A=\bigcup_{a\in A}\{a\}$ wich is a countable union of closed sets.
Also sets of the form  $\{x\}\times[0,1]$ or $[0,1]\times\{y\}$ are closed, hence are Borel measurable. Note that $C$ is a countable union of sets like this.
A: All given sets are Borel-measurable and therefore the Borel-measure coincides with the Lebesgue-measure.
