Step 1: Understanding the Concept:
This function is the Dirichlet function (a slight variation). We need to analyze its properties concerning Riemann integration. A function is Riemann integrable if its lower and upper Riemann integrals are equal. These integrals are defined using lower and upper sums (Darboux sums) over partitions of the interval.
Step 2: Key Formula or Approach:
Let \(P = \{x_0, x_1, \dots, x_n\}\) be a partition of \([a,b]\).
In any subinterval \([x_{i-1}, x_i]\), let:
- \(m_i = \inf\{f(x) \mid x \in [x_{i-1}, x_i]\}\)
- \(M_i = \sup\{f(x) \mid x \in [x_{i-1}, x_i]\}\)
The lower sum is \(L(P,f) = \sum_{i=1}^n m_i \Delta x_i\).
The upper sum is \(U(P,f) = \sum_{i=1}^n M_i \Delta x_i\).
The lower Riemann integral is \(\underline{\int_a^b} f(x)dx = \sup_P L(P,f)\).
The upper Riemann integral is \(\overline{\int_a^b} f(x)dx = \inf_P U(P,f)\).
The function is Riemann integrable if \(\underline{\int_a^b} f(x)dx = \overline{\int_a^b} f(x)dx\).
Step 3: Detailed Explanation:
The function is defined as:
\[ f(x) = \begin{cases} 1 & \text{if } x \text{ is rational} -1 & \text{if } x \text{ is irrational} \end{cases} \]
Let's analyze the properties on any interval \([a,b]\) where \(a<b\).
- (A) Boundedness: The function only takes values 1 and -1. So, \(|f(x)| \le 1\) for all \(x\). The function is bounded. Thus, statement (A) is false.
- (C) and (D) Riemann Integrability:
Consider any partition \(P\) of \([a,b]\) and any subinterval \([x_{i-1}, x_i]\).
Due to the density of both rational and irrational numbers in the reals, every such subinterval will contain both rational and irrational numbers.
Therefore, for every subinterval \([x_{i-1}, x_i]\):
- The supremum of \(f(x)\) is \(M_i = \sup\{1, -1\} = 1\).
- The infimum of \(f(x)\) is \(m_i = \inf\{1, -1\} = -1\).
Now, let's compute the upper and lower sums:
- Upper sum: \(U(P,f) = \sum_{i=1}^n M_i \Delta x_i = \sum_{i=1}^n 1 \cdot \Delta x_i = \sum_{i=1}^n \Delta x_i = b-a\).
- Lower sum: \(L(P,f) = \sum_{i=1}^n m_i \Delta x_i = \sum_{i=1}^n (-1) \cdot \Delta x_i = - \sum_{i=1}^n \Delta x_i = -(b-a)\).
Since these values are constant for any partition P, the upper and lower Riemann integrals are:
- Upper integral: \(\overline{\int_a^b} f(x)dx = \inf_P U(P,f) = b-a\).
- Lower integral: \(\underline{\int_a^b} f(x)dx = \sup_P L(P,f) = -(b-a)\).
Since \(b>a\), we have \(b-a \neq -(b-a)\).
The lower and upper Riemann integrals exist, but they are not equal. Therefore, the function is not Riemann integrable.
- This confirms that statement (C) is correct and statement (D) is incorrect. Statement (B) is also incorrect as the integrals exist.
Step 4: Final Answer:
For the given function, the lower and upper Riemann integrals exist but are not equal. This corresponds to option (C).