the function f(x)= { 0 , if x is irrational

{ 1 , if x is rational is :

a) continuous at x=1

b) discontinous only at 0

c)discontinous only at 0, 1

d) discontinous everywhere

The function f defined by f(irrational) = 0
and f(rational) = 1
is bounded but not integrable since its set of discontinuity points is not a zero set (since it is discontinuous everywhere).

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