I start by saying that my background on classical $L^p$ spaces and Lebesgue integration is not strong. There is mainly only one result that I keep in mind about Lebesgue integration and that is: if the Riemann integral exists, then the Lebesgue integral coincides with the Riemann one. There is certainly one question that comes up from this, which is: what happens when the Riemann integral is infinite? On the other hand, how can one evaluate a Lebesgue integral as infinite? All of this question came up from a specific question, which I present below:
Question. Show that the function $f(x) = \frac{5}{6} x^{-2/5}$ is not in $L^3 ( \, ]0,1[ \, ).$
In this post, for example, the answerer compares the function they're studying with a characteristic function (for which we know how to compute Lebesgue integrals explicity, since we just need to know the measure of the set in question). I haven't seen any other way of showing that a given Lebesgue integral is infinite and I don't think this same procedure can be applied to my specific question.
My work thus far. Obviously, I wish to prove that
$$ \int_0^1 |f(x)|^3 dx = +\infty. $$
It is clear that this happens if we are talking about Riemann integrals, but surely that's not enough to reach anywhere. So, I would like to ask for help /tips about on how to procede in this kind of scenarios.
Thanks for any help in advance.
A general way to check if a nonnegative function $f$ is in $L^p$ is to check its level sets. It's a theorem that a nonnegative function $f\in L^p$ if and only if $$ \sum_{n\in\mathbb Z}2^{np}|\underbrace{\{x:2^n < f(x)\le 2^{n+1}\}}_{\equiv E_n}| < \infty. $$ The rationale is that $f(x)^p \approx (2^n)^p$ exactly when $x\in \{2^n<f\le 2^{n+1}\}$, and these level sets are disjoint, so the Lebesgue integral $\int f^p \approx \sum_{n\in\mathbb Z}2^{np}|E_n|$ by the Monotone Convergence Theorem. (I am using $|E|$ to represent the Lebesgue measure of $E$.)
Coming to the example you asked about, $f(x) = x^{-2/5}$, so the level set $$ E_n = \{x\in (0,1]:2^n < f(x)\le 2^{n+1}\} $$ is an interval of length (Lebesgue measure) $\approx 2^{-\frac{5n}{2}}$ whenever $n \ge 0$, and they are empty when $n < 0$. Therefore, we just have to check the convergence/divergence of $$ \sum_{n=0}^\infty 2^{3n}2^{-5n/2}=\sum_{n=0}^\infty 2^{n/2}, $$ which diverges to $+\infty$. Hence $f\notin L^3((0,1])$. Based on this computation and the level sets criterion, this particular $f\in L^p$ if and only if $p < 5/2$.