Let f(x) = 0 if x is any rational number and f(x) = 1 if x is any irrational number. Show that f is not integrable on [0, 1].

Question

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Let f(x) = 0

 if x is any rational number and f(x) = 1

 if x is any irrational number.

Show that f is not integrable on [0, 1].

Answer 1

Let P : 0 = x₀ < x₁ < ••• < x_n = 1 be an arbitrary partition of [0,1]. Since each subinterval [x_i-1, x_i] contains a rational point r_i and an irrational point t_i, and f(r_i) = 0, f(t_i) = 1. Hence the lower Riemann sum L(f,P) = 0 and the upper Riemann sum U(f,P) = ∑[i = 1, n] Δx_i = 1 - 0 = 1. Since P was arbitrary, it follows that the upper Riemann integral of f is 1 and the lower Riemann integral of fis 0. Since the upper and lower Riemann integrals of f are not equal, f is not Riemann integrable on [0,1].