Quote Originally Posted by Jose27 View Post
I've been thinking about this but can't seem to make any progress. Can you prove the statement when instead of a countable family we ask it be finite? (Shame, I can't even prove this, not even for two!).

As an aside, if the sets are asked to be closed then the problem is trivial by Tietze's extension theorem.
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.