# Lebesgue vs. Riemann integrable function

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?)

• I presume you meant $\lambda > -1$ in the second? And yes, it exists as a Lebesgue integral for $\lambda >-1$. The catch with the Riemann integral for the $\lambda \in (-1,0)$ case is that the integrand is unbounded near zero. Commented Jun 6, 2013 at 5:52

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.
• @mathusiast I started with the assumption that the integral is finite, that is, the function is integrable for $\lambda \leq -1$. Since it is integrable, $t^\lambda \chi_{[\frac{1}{n},1]}$ is integrable and since the function is smaller than or equal to $t^{\lambda}$, the integral is at least not larger. But this becomes arbitrarily large. This contradicts the first assumption that was made, that the integral is finite.