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.