Math Problem Statement
Solution
The problem involves analyzing the function defined on the interval , given as follows:
\begin{cases} 1 & \text{if } x \in \mathbb{Q} \text{ (rational numbers)} \\ -1 & \text{if } x \in \mathbb{Q}^c \text{ (irrational numbers)} \end{cases}$$ You are asked to examine: 1. The integrability of the function $$f(x)$$. 2. The properties of the integral of $$f(x)$$, if it exists. --- ### Analysis: #### 1. **Examining Integrability (Riemann Integrability)**: A function is Riemann integrable on an interval if: - It is bounded on that interval. - The set of discontinuities has measure zero (i.e., the set of discontinuities does not "cover" any portion of the interval in terms of length). - **Boundedness**: The function $$f(x)$$ takes values $$1$$ and $$-1$$, so it is bounded on $$[0,1]$$. - **Discontinuities**: - The function is discontinuous at every point in $$[0,1]$$ because it jumps between $$1$$ (for rational $$x$$) and $$-1$$ (for irrational $$x$$). - The rational numbers ($$\mathbb{Q}$$) and irrational numbers ($$\mathbb{Q}^c$$) are densely interwoven in $$[0,1]$$. Hence, the discontinuities of $$f(x)$$ cover the entire interval. Thus, $$f(x)$$ is **not Riemann integrable** because the set of discontinuities is the entire interval $$[0,1]$$, which has non-zero measure. --- #### 2. **Lebesgue Integrability**: Under the Lebesgue approach, integrability depends on whether the absolute value of the function is integrable (i.e., its integral is finite). - The function $$f(x)$$ is measurable because both $$\mathbb{Q}$$ and $$\mathbb{Q}^c$$ are measurable sets. - The function's absolute value is constant ($$|f(x)| = 1$$), so its integral over $$[0,1]$$ is simply the length of the interval: $$\int_{[0,1]} |f(x)| dx = 1.$$ Thus, $$f(x)$$ is **Lebesgue integrable**. --- #### 3. **Value of the Integral**: - Since the function is $$1$$ on $$\mathbb{Q}$$ and $$-1$$ on $$\mathbb{Q}^c$$, the contributions from the two sets cancel each other out due to the symmetry of their densities. Therefore, the Lebesgue integral of $$f(x)$$ is: $$\int_{[0,1]} f(x) dx = 0.$$ --- ### Conclusion: 1. $$f(x)$$ is **not Riemann integrable**. 2. $$f(x)$$ is **Lebesgue integrable**, and its integral over $$[0,1]$$ is $$0$$. Would you like a deeper explanation of any part? Here are 5 related questions to expand understanding: 1. Why is Riemann integrability stricter than Lebesgue integrability? 2. What does "measure zero" mean in terms of discontinuities? 3. Can we generalize this result to similar functions on other intervals? 4. How does the density of rational and irrational numbers affect integrability? 5. What are the practical implications of a function being Lebesgue integrable but not Riemann integrable? **Tip:** For functions with dense discontinuities, consider switching to the Lebesgue framework for integration.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Riemann integrability
Lebesgue integrability
Properties of rational and irrational numbers
Set theory (measure zero)
Formulas
Riemann integrability conditions
Lebesgue integral formula ∫ |f(x)| dx
Theorems
A function is Riemann integrable if its set of discontinuities has measure zero.
Measurability and integrability in the Lebesgue sense.
Suitable Grade Level
Undergraduate Mathematics
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