Since our calculus lectures, we know that there are nowhere continuous functions (like the indicator function of the rationals). However, if we change this Dirichlet function on a set of measure zero, then we get a continuous function (the zero function). So my question ist
Does there exist a function f:R→R such that every function which equals f almost everywhere (with respect to the Lebesgue measure) is nowhere continuous?
Is it possible to choose f borel-measurable?
Answer
See my answer to this question for the construction of an Fσ set M⊂R such that 0<m(M∩I)<m(I) for every finite interval I, where m is the Lebesgue measure.
Let f be the indicator function of M. Clearly f is Borel-measurable. If g(x)=f(x) almost everywhere, then g−1(0) and g−1(1) are everywhere dense, whence g is nowhere continuous.
No comments:
Post a Comment