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. 
 A: 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
$$
\chi_{(0,a)}(x) = \begin{cases} 1 & \text{if } x\in(0,1), \\ 0 & \text{otherwise}. \end{cases}
$$
A: 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.
