While trying to learn the difference between Lebesgue and Riemann integrals, I came across the following example: $$\int_{0}^{1}t^\lambda\,\mathrm dt$$ What I know so far:
- only for $\lambda>0$ the integral exists as a Riemann integral,
- only for $\lambda>1$ it exists as an improper Riemann integral.
My question is: for which $\lambda$'s does it exist as a Lebesgue integral?
(I suspect it's the same as with the improper case since $|t^\lambda|=t^\lambda$, or is there more to it?)
For $\lambda > -1$, use Monotone Convergence Theorem on $t^\lambda \chi_{[\frac{1}{n},1]}$, where $\chi$ is the characteristic function.
For $\lambda \leq -1 $, assume that the integral is finite. Then it is finite for each $t^\lambda \chi_{[\frac{1}{n},1]}$. But this integral gets arbitrarily large as $n$ goes to infinity which is a contradiction.