The following is an exercise from Carothers' Real Analysis:

Show that $$\int_{1}^{\infty}\frac{1}{x}=\infty$$ (as a Lebesgue Integral).


Let $E=[1,\infty)$.

$\int_E f=\int f\cdot \chi_E=\sup\{\int\varphi:\varphi \text{ simple }, 0\leq \varphi\leq f\}\cdot \chi_E$

I'm not sure how to find out what $\sup\{\int\varphi:\varphi \text{ simple }, 0\leq \varphi\leq f\}$ is. (Maybe I can say something about $\sum_{n=1}^{\infty}\frac{1}{n}\cdot \chi_\mathbb{R}$, which diverges since it's the harmonic series?) I note that I can express $E=\bigcup_{n=1}^{\infty}[1,n)$, which is measurable since it is the union of measurable sets.

I'm pretty new to the Lebesgue integral so a hint would be preferred over a full solution if possible. Thanks.


To show that the Lebesgue integral of $x^{-1}$ is infinite, use a change of variables $x \mapsto 1/x$ and consider the sequence of simple functions

$$\phi_n = \sum_{j=1}^nj \chi_{((j+1)^{-1},j^{-1})}.$$

Note that $0 \leqslant \phi_n(x) \leqslant x^{-1}$ and

$$\int_{(0,1]}\phi_n = \sum_{j=1}^n j\left(\frac1{j}- \frac1{j+1}\right)=\sum_{j=1}^n \frac1{j+1}\to_{n \to \infty} +\infty.$$


$$\int_{[1,\infty)}x^{-1} = \int_{(0,1]}x^{-1} = +\infty$$

  • $\begingroup$ Thanks. Is there a particular reason you that you knew that choice for the simple function would work, or is this just from practice from doing a lot of these integrals? $\endgroup$ – Sujaan Kunalan Mar 22 '15 at 17:38
  • $\begingroup$ You're welcome. Originally, I misread your problem and started on the integral over $[0,1]$. However, I continued since this is equivalent to the integral over $[1, \infty)$. As an improper Riemann integral this, of course, is straightforward since we know the anti-derivative is $\ln x$. Working with the definition of the Lebesgue integral, I just took the first sequence of simple functions that came to mind -- leading to a partial sum of the harmonic series, since $\sum_{k=1}^n 1/k$ grows like $\ln n$. $\endgroup$ – RRL Mar 22 '15 at 18:35
  • 1
    $\begingroup$ Or use $\phi_n = \sum_{k=1}^n\frac{1}{k+1}\chi_{[k,k+1)}$ to show $\int_1^{\infty}x^{-1} = +\infty$, directly. $\endgroup$ – RRL Mar 22 '15 at 19:18
  • $\begingroup$ Would this be a sufficient proof:\begin{align*} &\int_1^{\infty}\frac{1}{x}\\ &=\int \sum_{k=1}^{\infty}\frac{1}{k+1}\chi_{(k,k+1]}\chi_{[1,\infty)}\\ &=\int \sum_{k=1}^{\infty}\frac{1}{k+1}\chi_{(k,k+1]\cap[1,\infty)}\\ &= \sum_{k=1}^{\infty}\frac{1}{k+1}\int\chi_{(k,k+1]\cap[1,\infty)}\\ &= \infty \text{ (since the harmonic series is divergent)} \end{align*} $\endgroup$ – Sujaan Kunalan Mar 22 '15 at 23:03
  • 1
    $\begingroup$ @SujaanKunalan: Technically you can't assume equality a priori in the second line (although both sides do diverge to $+\infty$). I would rather argue that with simple functions $\phi_n = \sum_{k=1}^n\frac{1}{k+1}\chi_{[k,k+1)}$ we have $\phi_n(x) < x^{-1}$ and $\sup \int \phi_n = \sup \sum_{k=1}^n 1/ (k+1) = +\infty$ so $\sup_{\phi \leqslant x^{-1}}\int \phi = +\infty$. Of course, this also follows from the divergence of the harmonic series. $\endgroup$ – RRL Mar 23 '15 at 0:02

Hint: Consider $f_n:x \to \frac{n}{\left\lceil nx\right\rceil}$. Since $\left\lceil nx\right\rceil\geq nx$, we have $\frac{1}{x}\geq f_n(x)$.


Fix $K>0$. Then there exists $N$ such that $\sum_2^N \frac1n>K$. Consider $$ f_K(x)=\sum_2^N\frac1n\,\chi_{[n-1,n]}. $$ Then $f_K(x)\leq f(x)$ for all $x$, and $\int f_K>K$. Then $$ \int_E f=\sup\{\int_E\varphi:\varphi \text{ simple }, 0\leq \varphi\leq f\}=\infty. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.