I have to prove that $$f(x)=\frac{1}{x^2ln(x)^3} $$ is not Lebesgue integrable.
I have tried doing this:
(USING THIS DEFINITION: $f \text{ (measurable function) Lebesgue integrable} \Leftrightarrow \int_{(0,+\infty)}|f| d\mu<\infty$)
$$\int_{(0,+\infty)}|f| d\mu=\int_{(0,+\infty)}\left|\frac{1}{x^2ln(x)^3}\right| d\mu=\int_{(0,1)}\frac{1}{x^2ln(x)^3} d\mu + \int_{(1,+\infty)}\frac{-1}{x^2ln(x)^3} d\mu$$
Then, $$\int_{(0,1)}\frac{1}{x^2ln(x)^3} d\mu=?=\int_0^1\frac{1}{x^2ln(x)^3} dx\geq \int_0^1\frac{1}{x^2} dx=+\infty $$
But, as we can see, I have proved that the functions Riemann's improper integral is divergent. Can I say that its Lebesgue integral is divergent too? I don't think so.
Apart from that, I don't know how to prove that $f$ is measurable. I've tried using that it is continuous, but it isn't when $x=1$. Thank you
The Riemann integral of a continuous function on close interval coincides with its Lebesgue integral.
$\int_a^{1-a} |f(x)| dx$ is a Riemann integral for each $a \in (0,\frac 1 2)$. And $\int_a^{1-a} |f(x)| dx \to \infty$ as $a$ decreases to $0$ since $|ln x| \leq |\ln a|$ on $(a,1-a)$ and $\int_a^{1-a} \frac 1 {x^{2}}dx\to \infty$ by direct calculation of the integral. Since $\int_0^{1} |f(x)| dx \geq \int_a^{1-a} |f(x)| dx$ we are done.
Hint for measurability: $\{x \neq 1: f(x) <a\}=\{x \in (0,1) \cup (1,\infty): \frac 1 {x (\ln x)^{3}} <a\} $ which is an open set by continuity. The function is not defined at $x=1$. If you define it say as $0$ at this point what happens to $\{x: f(x) <a\}$?