Let's say we have Borel $\sigma$-algebra for the interval $(0,1]$. Borel $\sigma$- algebra will have all singleton sets. Which implies it will contain all rationals and irrationals. But i doubt that there will not be any set in Borel $\sigma$-algebra which will contain all irrationals. Because we only have countable additivity axiom, and all irrationals are uncountable, and hence cant be build from taking unions of all singleton sets of irrationals. But since there will be a set which will contain all rationals, because its just union of all countable singleton sets of rationals. But if such a set exist then, our Borel $\sigma$; algebra must have its complement(a set with all its elements irrationals). Is such set exists(Containing all irrationals)??
Why such a question?? I am learning Probability theory where i found that professor assigned the measure to all rationals $=0$, and said that since irrationals are the complement of that, assigned measure $=1$ to them.So i got confused and posted a question. He is right in his way, because there is a set which includes all rationals, so its complement must be there.