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

Question

Let $A$ be a square matrix. If $A^2 = A$, then the matrix $A$ is called:

Answer 1

The key idea here is that the Cantor set $C$ has Lebesgue measure zero, so let's recompute the measure of $C$ from scratch and then see why the values of $f$ on $C$ never affect the integral.

The Cantor set is built by repeatedly removing the open middle third of each remaining interval. At step $1$ we remove a piece of length $\tfrac13$, at step $2$ we remove $2$ pieces of length $\tfrac19$ each, and at step $n$ we remove $2^{n-1}$ pieces of length $\tfrac{1}{3^n}$. The total length removed is the geometric series $\sum_{n=1}^{\infty} \tfrac{2^{n-1}}{3^n} = \tfrac13\cdot\dfrac{1}{1-\tfrac23}=1$. Since we started with an interval of length $1$ and removed a total length of $1$, the Cantor set has Lebesgue measure $0$.

Because $m(C)=0$, no matter how large $x^{2026}$ gets on $C$, the piece $\int_C x^{2026}\,dx=0$ (a bounded function integrated over a measure zero set is always $0$). Also, removing the measure zero set $C$ from $[0,\tfrac12]$ or $[\tfrac12,1]$ does not change the Lebesgue integral over those pieces, so we may integrate $\cos(\pi x)$ and $\sin(\pi x)$ over the full sub-intervals as ordinary integrals.

$\tfrac{2}{\pi}$: matches $\int_0^{1/2}\cos(\pi x)\,dx + \int_{1/2}^{1}\sin(\pi x)\,dx = \tfrac1\pi+\tfrac1\pi$.
$\tfrac1\pi$: this is only one of the two pieces, so it undercounts the total by half.
$\tfrac3\pi$: this would need an extra $\tfrac1\pi$ that has no source in the problem, an overcount.
$0$: this would only be right if the two pieces canceled out, but both $\int_0^{1/2}\cos(\pi x)\,dx$ and $\int_{1/2}^1\sin(\pi x)\,dx$ are positive, so they add instead of cancel.

Now compute the two pieces directly. First, $\int_0^{1/2}\cos(\pi x)\,dx = \left[\dfrac{\sin(\pi x)}{\pi}\right]_0^{1/2} = \dfrac{\sin(\pi/2)-\sin 0}{\pi} = \dfrac1\pi$. Second, $\int_{1/2}^{1}\sin(\pi x)\,dx = \left[-\dfrac{\cos(\pi x)}{\pi}\right]_{1/2}^{1} = -\dfrac{\cos\pi}{\pi}+\dfrac{\cos(\pi/2)}{\pi} = \dfrac1\pi+0=\dfrac1\pi$.

Let's summarize:

The Cantor set has measure zero, so the $x^{2026}$ branch of $f$ never contributes to the integral.
The integral reduces to $\int_0^{1/2}\cos(\pi x)\,dx+\int_{1/2}^1\sin(\pi x)\,dx=\tfrac1\pi+\tfrac1\pi=\tfrac2\pi$.

So the value of the Lebesgue integral of $f$ over $[0,1]$ is $\tfrac{2}{\pi}$.

\[ \boxed{\dfrac{2}{\pi}} \]

Answer 2

The key idea here is that the Cantor set $C$ has Lebesgue measure zero, so let's recompute the measure of $C$ from scratch and then see why the values of $f$ on $C$ never affect the integral.

The Cantor set is built by repeatedly removing the open middle third of each remaining interval. At step $1$ we remove a piece of length $\tfrac13$, at step $2$ we remove $2$ pieces of length $\tfrac19$ each, and at step $n$ we remove $2^{n-1}$ pieces of length $\tfrac{1}{3^n}$. The total length removed is the geometric series $\sum_{n=1}^{\infty} \tfrac{2^{n-1}}{3^n} = \tfrac13\cdot\dfrac{1}{1-\tfrac23}=1$. Since we started with an interval of length $1$ and removed a total length of $1$, the Cantor set has Lebesgue measure $0$.

Because $m(C)=0$, no matter how large $x^{2026}$ gets on $C$, the piece $\int_C x^{2026}\,dx=0$ (a bounded function integrated over a measure zero set is always $0$). Also, removing the measure zero set $C$ from $[0,\tfrac12]$ or $[\tfrac12,1]$ does not change the Lebesgue integral over those pieces, so we may integrate $\cos(\pi x)$ and $\sin(\pi x)$ over the full sub-intervals as ordinary integrals.

$\tfrac{2}{\pi}$: matches $\int_0^{1/2}\cos(\pi x)\,dx + \int_{1/2}^{1}\sin(\pi x)\,dx = \tfrac1\pi+\tfrac1\pi$.
$\tfrac1\pi$: this is only one of the two pieces, so it undercounts the total by half.
$\tfrac3\pi$: this would need an extra $\tfrac1\pi$ that has no source in the problem, an overcount.
$0$: this would only be right if the two pieces canceled out, but both $\int_0^{1/2}\cos(\pi x)\,dx$ and $\int_{1/2}^1\sin(\pi x)\,dx$ are positive, so they add instead of cancel.

Now compute the two pieces directly. First, $\int_0^{1/2}\cos(\pi x)\,dx = \left[\dfrac{\sin(\pi x)}{\pi}\right]_0^{1/2} = \dfrac{\sin(\pi/2)-\sin 0}{\pi} = \dfrac1\pi$. Second, $\int_{1/2}^{1}\sin(\pi x)\,dx = \left[-\dfrac{\cos(\pi x)}{\pi}\right]_{1/2}^{1} = -\dfrac{\cos\pi}{\pi}+\dfrac{\cos(\pi/2)}{\pi} = \dfrac1\pi+0=\dfrac1\pi$.

Let's summarize:

The Cantor set has measure zero, so the $x^{2026}$ branch of $f$ never contributes to the integral.
The integral reduces to $\int_0^{1/2}\cos(\pi x)\,dx+\int_{1/2}^1\sin(\pi x)\,dx=\tfrac1\pi+\tfrac1\pi=\tfrac2\pi$.

So the value of the Lebesgue integral of $f$ over $[0,1]$ is $\tfrac{2}{\pi}$.

\[ \boxed{\dfrac{2}{\pi}} \]

Show Hint

The Correct Option is A

Solution and Explanation

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