Define the function f : [0, 1] → R, { f(x) = 0 1 n if x irrational or x = 0, if x rational and x = m, with (m, n) = 1, n > 0. (a) (*) Prove that ƒ is Riemann integrable. Hint: Prove first that f is continuous at every irrational point and discontinuous at every rational point. Then use Theorem 11.33 (b) PMA page 323, and the fact that any countable subset of R is of measure zero (b) Find Riemann integrable functions f and g so that go f is defined on [0, 1], but is not Riemann integrable.
![Define the function f : [0, 1] → R,
f(x):
=
0
1
n
if x irrational or x =0,
if x rational and x = m, with (m, n) = 1, n > 0.
(a) (*) Prove that f is Riemann integrable.
Hint: Prove first that f is continuous at every irrational point and discontinuous
at every rational point. Then use Theorem 11.33 (b) PMA page 323, and the fact
that any countable subset of R is of measure zero
(b) Find Riemann integrable functions f and g so that go f is defined on [0, 1], but
is not Riemann integrable.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15ed467-90ec-4e60-afef-3d3f6119f74d%2F6562e8b8-e6b4-451c-9ea0-141ece4f9636%2F5oi8uif_processed.png&w=3840&q=75)
(a) To show that f is Riemann integrable on [0,1], we need to show that for any partition P of [0,1], the upper Darboux sum S*(f,P) and lower Darboux sum S(f,P) converge to the same limit L as the norm of the partition goes to zero.
Let P be any partition of [0,1]. We will show that S*(f,P) - S(f,P) ≤ ε for any ε > 0, which implies that f is integrable on [0,1].
Let ε > 0 be given. Since any countable subset of R is of measure zero, the set of rational numbers in [0,1] has measure zero.
Therefore, we can choose a partition P' of [0,1] such that the set of rational points in each subinterval of P' has measure less than ε/2.
Now, for each irrational point x in [0,1], we can find a subinterval of P' that contains x and has length less than ε/2. Since f is continuous at x, there exists a δ > 0 such that for any y in the subinterval of length δ around x, |f(y) - f(x)| < ε/2.
Therefore, for each subinterval of P', we can choose a point y in the subinterval that is either irrational or has a denominator greater than 1/δ. Then, we have
S*(f,P) - S(f,P) = Σ*(f(x_i) - f(y_i))(Δx_i) - Σ(f(x_i) - f(y_i))(Δx_i) ≤ (ε/2)(Σ(Δx_i)) + (1/n)(Σ(Δx_i)) = ε/2 + (1/n)(1-0) = ε/2 + 1/n
where x_i and y_i are any points in the i-th subinterval of P'. Note that the second term is bounded by 1/n because the denominators of the rational points in each subinterval are greater than 1/δ.
Since ε can be arbitrarily small, we have shown that S*(f,P) - S(f,P) converges to 0 as the norm of the partition goes to zero, and hence f is Riemann integrable on [0,1].
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