I'm stuck with this problem on Lebesgue Integration and measure theory (I'm a newbie in this subject). I have to do the following integral:
$$ \lim_{n\rightarrow\infty}{\int_{0}^{\pi}\left(\sin{x} \right)^\frac{1}{n}\left| \log{x} \right|dx } $$
I know, in order to prove that I can enter the limit operator into the integral one, it must satisfy the conditions of Lebesgue's Dominated Convergence Theorem, which are:
- The sequence of functions $f_n$ must be measurable
- The sequence of functions $f_n$ must converge a.e to a function $f$
- There must exist a function $g$ $\mu$-integrable such that $\forall\ n\in \mathbb{Z}^+$, $|f_n|\le g$
Here is my attempt:
- To show that the sequence of $f_n=\left(\sin{x} \right)^\frac{1}{n}\left| \log{x} \right|$ is measurable I define the domain $X = (0, \pi] $ for the $f_n$ and I try to apply that if a function $y$ is continuous in $X$ then it is measurable. Is this way correct?
- Such limit equals to $f=\left| \log{x} \right| $, although I don't actually know if it is uniform or pointwise. Is it important?
- In the third one, can I take $g=f=\left| \log{x} \right|$ as the dominate function?
I believe I can swap Limit and integral operators, but cannot show it mathematically. Any suggestions?
Thank you all
PD: I reckon this is not a duplicate question of Calculate $\lim\limits_{n\rightarrow\infty} \int_0^\pi \sin(x)^{1/n}|\log(x)| dx$ because there is not showed formaly what i want. It doesn't matter that the function is the same, what matters to me is the theory behind