MHB Compute the integral of the Cantor-Lebesgue function ∫_0^1〖F(x)〗 dm .

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The integral of the Cantor-Lebesgue function ∫_0^1 F(x) dm is definitively 0. This result arises from the nature of the Cantor-Lebesgue function as a singular function, which possesses a zero Lebesgue measure. To compute this integral, one applies the definition of the Lebesgue integral: ∫_0^1 F(x) dm = lim_{n→∞} ∑_{i=1}^n F(x_i)Δm_i, where x_i are points in the interval [0,1] and Δm_i represents the Lebesgue measure of the intervals. Given that the function is singular, all intervals between the points have a measure of 0, confirming that the integral evaluates to 0.

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How should I compute the integral of the Cantor-Lebesgue function ∫_0^1〖F(x)〗 dm?
 
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The integral of the Cantor-Lebesgue function is 0. This is due to the fact that the Cantor-Lebesgue function is a singular function and thus has a zero Lebesgue measure. To compute this integral, one can use the definition of the Lebesgue integral: ∫_0^1〖F(x)〗 dm = lim_{n→∞} ∑_i=1^n F(x_i)Δm_iwhere x_i are points in the interval [0,1] and Δm_i is the Lebesgue measure of the interval between them. Since the function is singular, all of the intervals between the points will have a measure of 0, so the integral will be 0.
 

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