Show that the Dirichlet function f defined on [0, 1] by f(x) = { 1 if x is rational, 0 if x is irrational is not Riemann integrable on [0, 1].
Let f : [-1, 1] → R be a bounded function, and let ̑ : R → R be defined by: ̑(x) = { 0 if x < 0, 1/2 if x = 0, 1 if x > 0. Show that f ∈ R(̑) on [-1, 1] if and only if f is continuous at x = 0, and in this case, find the value of ∫_{-1}^1 f d̑.