Answer to Question #192369 in Calculus for Nikhil Rawat

Question

The function,f: [-1,1]×[2,1] → R, defined by

f(x,y)= { x ; y is rational

{ 0 ; y is irrationa

is integrable or not.

True or false with full explanation

Answer 1

A bounded function on a compact interval [a, b] is Riemann integrable if and only if it is continuous almost everywhere (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure).

In our case, the function is discontinuous at every irrational point. Since the set of irrational points on a segment has the cardinality of the continuum, the function is not integrable.

Answer: function is not integrable.