How does one show that $\chi_{[1, \infty)}1/x$ is not (Lebesgue) integrable?
What I could think of is as follows:
Letting $f(x)=1/x$ (defined for $x\geq 1$), define $ f_n(x)=f\chi_{[1, n)}(x). $
Each $f_n$ is, therefore, Riemann integrable on $[1, n)$ with value $\ln n$, hence integrable there. As $0\leq f_n\nearrow f$ on $[1, \infty)$, the monotone increasing theorem says $ \int_{[1, \infty)}f_n\nearrow\int_{[1, \infty)} f $ and so $\int_{[1, \infty)} f=\infty$ since $\ln n\nearrow\infty$.
Is there a more obvious reason why the given integral isn't finite? It seems that my method needs quite some modification if we go to $n$-dimensional integrals of $ f(x)=\frac{1}{|x|}\chi_{|x|>1}. $