hello,Jose27,thank you for your reply.If {Ek} is finite,wo can let E to be all the rational numbers in R1.Suppose E={qn}n=1,2,...,and let f(qn)=n,n=1,2,...
Then we have
there doesn't exist finite Lebesgue measurable subsets {Ek}k=1,2,...
of E
such that
(1)the untion of Ek is E;
(2)the restriction of f to every set Ek is continuous on Ek.