Step 1: Understanding the Concept:
This question tests fundamental theorems regarding the conditions for Riemann integrability of a function. We need to identify which of the given implications is false.
Step 2: Key Formula or Approach:
We will analyze each statement based on standard theorems from Riemann integration theory.
- A function continuous on \([a,b]\) is Riemann integrable on \([a,b]\).
- A function monotonic on \([a,b]\) is Riemann integrable on \([a,b]\).
- The relationship between the integrability of \(f\) and \(|f|\) is a one-way implication.
- The relationship between the integrability of \(f\) and \(f^2\) is also a one-way implication.
Step 3: Detailed Explanation:
Let's evaluate each statement:
- (A) Every continuous function is integrable. This is a fundamental theorem of calculus. A function that is continuous on a closed and bounded interval \([a,b]\) is guaranteed to be Riemann integrable on that interval. So, this statement is CORRECT.
- (B) If f is monotonic on [a,b] then f is integrable in [a,b]. This is another fundamental theorem. Any function that is monotonic (either non-decreasing or non-increasing) on a closed and bounded interval is Riemann integrable. So, this statement is CORRECT.
- (D) If f is integrable on [a,b] then \(f^2\) is also integrable on [a,b]. This is a known property of Riemann integrable functions. If \(f\) is integrable, then \(f^2\) is also integrable. This can be proven by showing that the set of discontinuities of \(f^2\) has measure zero if the set of discontinuities of \(f\) has measure zero. So, this statement is CORRECT.
- (C) If \(|f|\) is integrable on [a,b] then f is integrable on [a,b]. This statement is the converse of another property. The property is "if \(f\) is integrable, then \(|f|\) is integrable". The converse, however, is not true. We need to find a counterexample.
Consider the Dirichlet-like function from Q71 on an interval, say \([0,1]\):
\[ f(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \cap [0,1] -1 & \text{if } x \in (\mathbb{R}-\mathbb{Q}) \cap [0,1] \end{cases} \]
As shown previously, this function \(f\) is not Riemann integrable.
However, consider \(|f(x)|\).
\[ |f(x)| = \begin{cases} |1| & \text{if } x \in \mathbb{Q} \cap [0,1] |-1| & \text{if } x \in (\mathbb{R}-\mathbb{Q}) \cap [0,1] \end{cases} = 1 \]
So, \(|f(x)| = 1\) for all \(x \in [0,1]\). This is a constant function, which is continuous and therefore Riemann integrable.
We have found a function \(f\) such that \(|f|\) is integrable, but \(f\) itself is not. Therefore, the implication in statement (C) is false.
Step 4: Final Answer:
The statement "If \(|f|\) is integrable on [a,b] then f is integrable on [a,b]" is NOT CORRECT. This corresponds to option (C).