How to do Lebesgue Integration

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 $f$. –  AD. Jun 10 '12 at 6:28
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 I am asked to show that it is Lebesgue integrable on the interval (0,1) and then to evaluate it. The function is bounded by 0 & 1. So does this automatically mean it is Lebesgue Integrable? When I evaluated the function on the interval from (0,1) I got $\int_0^1 f(x) =1$ am I on the right track? –  Francisco Nov 30 '11 at 6:02
The integration is off, $1$ can't be the answer, our function is $>1$. You lost a constant somewhere when you were integrating. –  André Nicolas Nov 30 '11 at 7:18
Francisco: it seems you are not asked to compute the value of the integral but to show that it exists. Hence, what you need is a result in your lecture notes saying that a function satisfying such and such property is Lebesgue integrable. Amongst these propertties might be that the Riemann integral on every interval $(\varepsilon,1)$ exists, and other similar things. So, what have you got in your lecture notes which looks like this? –  Did Nov 30 '11 at 7:37
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 en.wikipedia.org/wiki/… ) –  tomcuchta Nov 30 '11 at 18:47