TITLE: $f(x)=1/x$ over $[1, \infty)$ is not Lebesgue integrable
QUESTION [6 upvotes]: 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 
$$
f(x)=\frac{1}{|x|}\chi_{|x|>1}.
$$

REPLY [2 votes]: So we have that $\int f$ is the supremum of the integrals of simple functions of finite support no more than $f$. Let $\phi_{n} = \sum_{k = 1}^{n} \chi_{[k, k + 1)} / (k + 1)$. Then $\phi_{n}$ has measure $n$ and integral $\sum_{k = 1}^{n} 1/(k + 1)$. Thus $\int f \geq \sum_{k = 1}^{n} 1/(k + 1)$ for all $n$, so we also have $\int f \geq \sum_{k = 1}^{\infty} 1/(k + 1) = \infty$.