Question:
Let \( C \) denote the Cantor set and \( f: [0,1] \to \mathbb{R} \) be defined as follows:
\[ f(x) = \begin{cases} x^{2026} & \text{for } x \in C \\ \cos(\pi x) & \text{for } x \in \left[0, \dfrac{1}{2}\right] \setminus C \\ \sin(\pi x) & \text{for } x \in \left[\dfrac{1}{2}, 1\right] \setminus C \end{cases} \]
The value of the Lebesgue integral of \( f(x) \) over the interval \( [0,1] \) is equal to
Let \( C \) denote the Cantor set and \( f: [0,1] \to \mathbb{R} \) be defined as follows:
\[ f(x) = \begin{cases} x^{2026} & \text{for } x \in C \\ \cos(\pi x) & \text{for } x \in \left[0, \dfrac{1}{2}\right] \setminus C \\ \sin(\pi x) & \text{for } x \in \left[\dfrac{1}{2}, 1\right] \setminus C \end{cases} \]
The value of the Lebesgue integral of \( f(x) \) over the interval \( [0,1] \) is equal to
\[ f(x) = \begin{cases} x^{2026} & \text{for } x \in C \\ \cos(\pi x) & \text{for } x \in \left[0, \dfrac{1}{2}\right] \setminus C \\ \sin(\pi x) & \text{for } x \in \left[\dfrac{1}{2}, 1\right] \setminus C \end{cases} \]
The value of the Lebesgue integral of \( f(x) \) over the interval \( [0,1] \) is equal to
Show Hint
The Cantor set has Lebesgue measure zero, so its branch of f never affects the value of the integral.
Updated On: Jul 21, 2026
- \( \dfrac{2}{\pi} \)
- \( \dfrac{1}{\pi} \)
- \( \dfrac{3}{\pi} \)
- \( 0 \)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question.
The function \( f \) is defined piecewise, using one rule on the Cantor set \( C \), one rule on \( [0,\tfrac12]\setminus C \), and one rule on \( [\tfrac12,1]\setminus C \). We need the Lebesgue integral of \( f \) over \( [0,1] \).
Step 2: Key Formula or Approach.
The Cantor set \( C \) has Lebesgue measure zero. A key fact about the Lebesgue integral is that a bounded function integrated over a set of measure zero always gives \( 0 \), and changing a function's values on a measure zero set never changes its integral over a larger set.
Step 3: Detailed Explanation.
Since \( m(C) = 0 \), the contribution of the branch \( x^{2026} \) on \( C \) to the integral is
\[ \int_C x^{2026}\, dx = 0 \]
no matter how the values \( x^{2026} \) behave on \( C \).
Also, since \( C \) has measure zero, removing it from \( [0,\tfrac12] \) or from \( [\tfrac12,1] \) does not change the measure of those pieces, so the Lebesgue integral over \( [0,\tfrac12]\setminus C \) equals the ordinary integral of \( \cos(\pi x) \) over the full interval \( [0,\tfrac12] \), and similarly for \( \sin(\pi x) \) over \( [\tfrac12,1] \). So
\[ \int_0^1 f(x)\,dx = \int_0^{1/2} \cos(\pi x)\, dx + \int_{1/2}^{1} \sin(\pi x)\, dx \]
Compute the first piece:
\[ \int_0^{1/2} \cos(\pi x)\, dx = \left[\frac{\sin(\pi x)}{\pi}\right]_0^{1/2} = \frac{\sin(\pi/2)-\sin 0}{\pi} = \frac{1}{\pi} \]
Compute the second piece:
\[ \int_{1/2}^{1} \sin(\pi x)\, dx = \left[-\frac{\cos(\pi x)}{\pi}\right]_{1/2}^{1} = -\frac{\cos\pi}{\pi}+\frac{\cos(\pi/2)}{\pi} = \frac{1}{\pi}+0 = \frac{1}{\pi} \]
Adding the two pieces gives \( \tfrac1\pi + \tfrac1\pi = \tfrac2\pi \).
Option (B), \( \tfrac1\pi \), only counts one of the two pieces. Option (C), \( \tfrac3\pi \), adds an extra \( \tfrac1\pi \) that has no source in the problem. Option (D), \( 0 \), would only hold if the two pieces canceled, but both pieces are positive and add up instead.
Step 4: Final Answer.
The Lebesgue integral of \( f \) over \( [0,1] \) equals \( \tfrac2\pi \).
\[ \boxed{\dfrac{2}{\pi}} \]
The function \( f \) is defined piecewise, using one rule on the Cantor set \( C \), one rule on \( [0,\tfrac12]\setminus C \), and one rule on \( [\tfrac12,1]\setminus C \). We need the Lebesgue integral of \( f \) over \( [0,1] \).
Step 2: Key Formula or Approach.
The Cantor set \( C \) has Lebesgue measure zero. A key fact about the Lebesgue integral is that a bounded function integrated over a set of measure zero always gives \( 0 \), and changing a function's values on a measure zero set never changes its integral over a larger set.
Step 3: Detailed Explanation.
Since \( m(C) = 0 \), the contribution of the branch \( x^{2026} \) on \( C \) to the integral is
\[ \int_C x^{2026}\, dx = 0 \]
no matter how the values \( x^{2026} \) behave on \( C \).
Also, since \( C \) has measure zero, removing it from \( [0,\tfrac12] \) or from \( [\tfrac12,1] \) does not change the measure of those pieces, so the Lebesgue integral over \( [0,\tfrac12]\setminus C \) equals the ordinary integral of \( \cos(\pi x) \) over the full interval \( [0,\tfrac12] \), and similarly for \( \sin(\pi x) \) over \( [\tfrac12,1] \). So
\[ \int_0^1 f(x)\,dx = \int_0^{1/2} \cos(\pi x)\, dx + \int_{1/2}^{1} \sin(\pi x)\, dx \]
Compute the first piece:
\[ \int_0^{1/2} \cos(\pi x)\, dx = \left[\frac{\sin(\pi x)}{\pi}\right]_0^{1/2} = \frac{\sin(\pi/2)-\sin 0}{\pi} = \frac{1}{\pi} \]
Compute the second piece:
\[ \int_{1/2}^{1} \sin(\pi x)\, dx = \left[-\frac{\cos(\pi x)}{\pi}\right]_{1/2}^{1} = -\frac{\cos\pi}{\pi}+\frac{\cos(\pi/2)}{\pi} = \frac{1}{\pi}+0 = \frac{1}{\pi} \]
Adding the two pieces gives \( \tfrac1\pi + \tfrac1\pi = \tfrac2\pi \).
Option (B), \( \tfrac1\pi \), only counts one of the two pieces. Option (C), \( \tfrac3\pi \), adds an extra \( \tfrac1\pi \) that has no source in the problem. Option (D), \( 0 \), would only hold if the two pieces canceled, but both pieces are positive and add up instead.
Step 4: Final Answer.
The Lebesgue integral of \( f \) over \( [0,1] \) equals \( \tfrac2\pi \).
\[ \boxed{\dfrac{2}{\pi}} \]
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