Discussion
Let C denote the Cantor set and \(f : [0,1] \to \mathbb{R}\) be defined as follows:
\[ f(x) = \begin{cases} x^{2026} & \text{for } x \in C \\ \cos(\pi x) & \text{for } x \in [0, \frac{1}{2}] \setminus C \\ \sin(\pi x) & \text{for } x \in [\frac{1}{2}, 1] \setminus C \end{cases} \]
The value of the Lebesgue integral of \(f(x)\) over the interval [0,1] is equal to
Answer: A
The Lebesgue integral equals \(\frac{2}{\pi}\).
The Cantor set has Lebesgue measure zero, so the values of f on it, including \(x^{2026}\), do not affect the integral. Off the Cantor set f is \(\cos \pi x\) on \([0, \frac{1}{2}]\) and \(\sin \pi x\) on \([\frac{1}{2}, 1]\):
\[ \begin{aligned} \int_0^{1/2} \cos \pi x\, dx &= \frac{1}{\pi} \\ \int_{1/2}^1 \sin \pi x\, dx &= \frac{1}{\pi} \end{aligned} \]
The sum is \(\frac{2}{\pi}\). \(\frac{1}{\pi}\) counts only one of the two pieces, 0 would follow only if the whole interval were negligible, and \(\frac{3}{\pi}\) wrongly adds a contribution from the Cantor set.
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