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How can you prove that the following is Lebesgue Integrable on $(0,1)$?

$ f(x)= \frac{1}{\sqrt[3]{1-x}} $

I don't know how to approach it with Lebesgue. Any help would be appreciated.

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The function is everywhere non-negative, and it's continuous except at one point, so saying it's Lebesgue-integrable is the same as saying that the integral is less than $\infty$. The easiest way to find the integral may be to treat it as $\lim\limits_{a\;\uparrow\; 1} \int_0^a$, but that's not how the Lebesgue integral is defined. So the question would be whether $\lim\limits_{a\;\uparrow\; 1} \int_0^a$ is the same as the integral you get from Lebesgue's definition. And that can come from the monotone convergence theorem applied to $f\cdot \chi_{(0,a)}$ where $\chi_{(0,a)}$ is the indicator function $ \chi_{(0,a)}(x) = \begin{cases} 1 & \text{if } x\in(0,1), \\ 0 & \text{otherwise}. \end{cases} $

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    Note that the above argument works fine for all positive functions $f$.2012-06-10
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It is a theorem that (in the sense of Lebesgue integrals) that $ \lim_{t\to b} \int^t_a f(x) dx = \int^b_a f(x) dx $ and it is also known that if a Riemann integral is defined, then so is the Lebesgue integral and they share the same value. Thus, improper Riemann integrals are just regular Lebesgue integrals and they share the same value.

So since you can show your integral is improperly Riemann integrable, you also know it is Lebesgue integrable.

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    You cannot conclude that since the integral is improper Riemann integrable that it is also Lesbesgue integrable. The theorem relating Lebesgue and Riemann integrals you are referring to requires that the function be bounded on a closed integral (see Royden (3rd edition) Chapter 4, Section 2, Proposition 4, pg. 81). In fact the integral $\displaystyle\int_0^{\infty} \frac{\sin(x)}{x} dx$ converges as an improper Riemann integral but not as a Lebesgue integral (see http://en.wikipedia.org/wiki/Improper_integral#Improper_Riemann_integrals_and_Lebesgue_integrals )2011-11-30