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 $$ f(x)=\frac{1}{|x|}\chi_{|x|>1}. $$

  • 1
    $\begingroup$ Hint: switch to polar coordinates $\endgroup$ – Chris Janjigian Oct 26 '12 at 3:55

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 $\int_{[1,+\infty)}1/|x|dx=+\infty$.

  • $\begingroup$ It's not strictly necessary to use MCT, which can simplify the prerequisites for the proof. All you need is, for each $N$, a simple function $g_N$ of integral greater than $N$ with $0 < g_N < 1/x$. That is enough to ensure the supremum of the integrals of simple functions less that $1/x$ is $\infty$. $\endgroup$ – Carl Mummert Mar 27 '13 at 12:01
  • $\begingroup$ @CarlMummert Apparently, the question has not been edited...So I guess I have not read it carefully, since I simply reapeated the OP's argument...I'll edit. $\endgroup$ – Julien Mar 27 '13 at 12:09

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 $\int f \geq \sum_{k = 1}^{\infty} 1/(k + 1) = \infty$.


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: $1 < f + \dfrac 1f$.

  • $\begingroup$ What do you mean by an infinite measure space, do you mean $\mu X = \infty$? How can $\int f$ be infinite if $f \in L_1(\mu)$? $\endgroup$ – copper.hat Oct 26 '12 at 3:46
  • $\begingroup$ Yes. $\mu(X) = \infty$. Note that $\int g\,d\mu$ makes sense for any $\mu$-measurable function $g: X \to [0,\infty]$ though of course the integral might be $+\infty$. $\endgroup$ – kahen Oct 26 '12 at 3:52
  • $\begingroup$ Yes, I realize that, but the notation $f \in L_1(\mu)$ means $\int |f| < \infty$. So how can the proposition be true? $\endgroup$ – copper.hat Oct 26 '12 at 3:53
  • $\begingroup$ D'oh. Typo. It should of course be $\frac 1f$. $\endgroup$ – kahen Oct 26 '12 at 3:54
  • $\begingroup$ @kahen $\frac{1}{x^2}$ on $[1,\infty)$ is a counterexample to this claim, no? $x^2 > 0$ on $[1,\infty)$ but $\frac{1}{x^2}$ is integrable or am I missing something? $\endgroup$ – Chris Janjigian Oct 26 '12 at 3:55

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.


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