If f doesn't keep the sign, then because the amplitude of f is small, the amplitude of |f| will not be large, but it should be smaller, because after taking the absolute value, it all runs to the edge of the number axis. Therefore, the small amplitude range of f is maintained at |f|. Therefore, the length of the interval with large amplitude of |f| will not exceed f, thus being controlled. The integrability of |f| is given by the integrability of F.
And the integrability of f 2 is similar. As long as there is this feeling, the square of f is even smaller where f is small. Because f is integrable and bounded, no matter how big it is, it will not exceed a constant, so don't worry about it.
Extended data:
Functions are related to inequalities and equations (elementary functions). Let the function value be equal to zero. From a geometric point of view, the value of the corresponding independent variable is the abscissa of the intersection of the image and the X axis.
From the algebraic point of view, the corresponding independent variable is the solution of the equation. In addition, replacing "=" in the expression of a function (except a function without expression) with "",and then replacing "y" with other algebraic expressions, the function becomes an inequality, and the value range of the independent variable can be found.
It is based on Lebesgue measure theory. This integral can deal with bounded and unbounded functions, and functions can also be defined on more general point sets. More importantly, it provides a more extensive and effective convergence theorem than Riemann integral. Therefore, the application of Lebesgue integral is more extensive, especially for the in-depth study of probability theory and mathematical statistics.
Baidu Encyclopedia-Integrable Function
Answered on 20 19-06-24.
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Proof: when f? (x) When it is integrable on [a, b], |f(x)| can also be integrable on [a, b].
If the Lebesgue integrability is known, where the integrability of f gives that f is at least a measurable function instead of 0, then 1/f is also a measurable function. So we define its integral. Because the absolute value of f is greater than a fixed number, its derivative is bounded. So the integral will not be infinite, so Leberg can be integrated. Of course, Lebesgue integrable and Riemann integrable are consistent in this sense.
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It is proved that f(x) can be integrated in (a, b), and its absolute value and square can also be integrated.
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If f(x) is integrable in [a, b], then its absolute value can also be integrable in [a, b]. How to prove it or is there a counterexample?
If integrability means Riemann integrability, the conclusion is correct; If it refers to generalized integrability, the conclusion is negative. The proof of Riemannian integrability needs the first or second necessary and sufficient condition of integrability. It is simpler to prove with the second necessary and sufficient condition: the necessary and sufficient condition of f integrability is that there is a division of [a, b] P for any given e>0: a=x0.
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