Question

Define f: [0,1] → [0,1] by
\(f(x) = \begin{cases} 1 & \text{if } x =0 \\ \frac{1}{n} & \text{if } x=\frac{m}{n}\ \text{for some}\ m, n \isin \N\ \text{with}\ m\le n\ \text{and gcd(m, n)} = 1, \\ 0 & \text{if} x\isin[0,1]\ \text{is irrational}\end{cases}\)
and define g: [0,1]→ [0,1] by
\(g(n) = \begin{cases} 0 & \text{if } x=0 \\ 1 & \text{if } x\isin(0,1]. \end{cases}\)
Then which of the following is/are true?

• A. f is Riemann integrable on [0,1].
• B. g is Riemann integrable on [0,1].
• C. The composite function f○g is Riemann integrable on [0,1].
• D. The composite function g○f is Riemann integrable on [0,1].

Answer 1

The Correct Option is A, B, C

Let's analyze each part of the problem to determine the Riemann integrability of the functions \(f\), \(g\), and their compositions \(f \circ g\) and \(g \circ f\) over the interval \([0,1]\). 

1. Function \(f\):
   • \(f(0) = 1\).
   • \(f(x) = \frac{1}{n}\) if \(x = \frac{m}{n}\), where \(m\), \(n \in \mathbb{N}\), \(m \leq n\) and \(\gcd(m, n) = 1\).
   • \(f(x) = 0\) if \(x\) is irrational.
2. Function \(g\):
   • \(g(0) = 0\).
   • \(g(x) = 1\) for \(x \in (0,1]\).
3. Composite Function \(f \circ g\):
   • For \(x = 0\), \(f(g(0)) = f(0) = 1\).
   • For \(x \in (0,1]\), \(f(g(x)) = f(1) = \frac{1}{1} = 1\).
4. Composite Function \(g \circ f\):
   • For \(x = 0\), \(g(f(0)) = g(1) = 1\).
   • For rational \(x = \frac{m}{n}\), \(g(f(x)) = g\left(\frac{1}{n}\right) = 1\).
   • For irrational \(x\), \(g(f(x)) = g(0) = 0\).

Based on the analysis, the following statements are true:

• \(f\) is Riemann integrable on \([0,1]\).
• \(g\) is Riemann integrable on \([0,1]\).
• The composite function \(f \circ g\) is Riemann integrable on \([0,1]\).