$f(x)=1/x$ over $[1, \infty)$ is not Lebesgue integrable 
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 
$$
f(x)=\frac{1}{|x|}\chi_{|x|>1}.
$$
 A: By definition, $\int_{[1,+\infty)}|f(x)|dx=\int_{[1,+\infty)}1/|x|dx$ is the sup, over all simple functions $0\leq s \leq |f|$, of the $\int_{[1,+\infty)}s(x)dx$. Now for each $N\geq 1$. Observe that
$$
0\leq s_N(x):=\sum_{n=1}^N \frac{1}{n+1} 1_{(n,n+1)}(x)\leq\frac{1}{|x|}\qquad\forall x\geq 1.
$$
Hence
$$
\int_{[1,+\infty)}1/|x|dx\geq \int_{[1,+\infty)}s_N(x)dx=\sum_{n=1}^N
\frac{1}{n+1}\longrightarrow +\infty$$
as the harmonic series is well-known to diverge (which can be proved elementarily, without resorting to the integral).
This proves $\int_{[1,+\infty)}1/|x|dx=+\infty$.
A: 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 $\int f \geq \sum_{k = 1}^{\infty} 1/(k + 1) = \infty$.
A: For additional insight, you should try to prove the following:

Proposition. If $(X,\mu)$ is an infinite measure space and $f \in \mathcal L_1(X,\mu)$ is positive, then $\displaystyle \int \frac1f\,d\mu = \infty$.

Hint: $1 < f + \dfrac 1f$.
A: Here's an alternative to using polar coordinates in $\mathbb{R}^n$:
If $x \neq 0$, we have $\frac{1}{\sqrt{n} \|x\|_\infty} \leq \frac{1}{\|x\|_2}$. Then $\frac{1}{\sqrt{n}} \int_{\|x\|_\infty \geq 1} \frac{1}{\|x\|_\infty} \leq \int_{\|x\|_\infty \geq 1} \frac{1}{\|x\|_2} \leq \int_{\|x\|_2 \geq 1} \frac{1}{\|x\|_2}$.
So we will consider $\int_{\|x\|_\infty \geq 1} \frac{1}{\|x\|_\infty}$ instead.
Let $B_k = B_\infty(0, k)$ and note that $m B_k = (2k)^n$ and $\frac{1}{\|x\|_\infty} > \frac{1}{k}$ on $B_k$. Then we have the lower bound
$$ \sum_{k=2}^K \frac{1}{k} m(B_k\setminus B_{k-1}) = 2^n\sum_{k=2}^K \frac{1}{k}(k^n-(k-1)^n) \leq \int_{\|x\|_\infty \geq 1} \frac{1}{\|x\|_\infty}$$
Since $k^n-(k-1)^n \geq 1$ for all $k\geq 1$, we have
$$2^n\sum_{k=2}^K \frac{1}{k} \leq \int_{\|x\|_\infty \geq 1} \frac{1}{\|x\|_\infty}$$
Non-integrability follows from this.
