Explain to me this example that shows $L^1([a,b])$ is not closed under multiplication.

Question

This is an exercise of my course of Measure and Integration.

The original question

Give an example that $f$ is Lebesgue integrable but $f^2$ is not.

What I found:

This is an example: $f:(0,1)\rightarrow\mathbb{R}$ such that $f(x)=1/\sqrt{x}$.

What I Know: By the Chebichevy's inequality we have that for each $\lambda>0$ $$ \infty>m(\{x\in(0,1): f(x)=1/\sqrt{x}\leq\lambda \})\geq\frac{1}{\lambda}\int_{(0,1)}f $$ so $f$ is integrable. ( $\int |f|<\infty$ ).

What I don't understand:

Why $f^2=1/x$ is not Lebesgue integrable? Furthermore give me another example for a function that is not Lebesgue integrable and tell me why.

Answer 1

$f(x)=\frac{1}{\sqrt{x}}$, $x \in (0,1)$. But I'm sure this question is a duplicate.