For each n ∈ N, define fn : [0, 1] → R byfn(x) =(xn cos(1/x) if x > 0,0 if x = 0.11. Prove that fn ∈ R[0, 1] for all n ∈ N

To prove that fn ∈ R[0, 1] for all n ∈ N, we need to show that fn is Riemann integrable on the interval [0, 1] for all natural numbers n.

Step 1: Define the function fn

For each natural number n, we define a function fn : [0, 1] → R as follows:

fn(x) = {xn cos(1/x) if x > 0, 0 if x = 0}

Step 2: Show that fn is bounded

The function fn is clearly bounded on the interval [0, 1] because for all x in [0, 1], |fn(x)| ≤ |xn| ≤ 1.

Step 3: Show that fn has a finite number of discontinuities

The function fn is continuous on the interval (0, 1] because it is a product of continuous functions on this interval. At x = 0, the function fn is defined to be 0, so it is also continuous at x = 0. Therefore, fn is continuous on the entire interval [0, 1] and has no discontinuities.

Step 4: Apply the Riemann integrability criterion

A function is Riemann integrable on an interval [a, b] if and only if it is bounded and has a finite number of discontinuities. Since we have shown that fn is bounded and has no discontinuities on the interval [0, 1], we can conclude that fn is Riemann integrable on this interval.

Therefore, fn ∈ R[0, 1] for all n ∈ N.