Lebesgue Integral: $\int_1^{\infty}\frac{1}{x}$

The following is an exercise from Carothers' Real Analysis:

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

Attempt:

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

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

Whence,

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

• 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? – Sujaan Kunalan Mar 22 '15 at 17:38
• 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 $\ln n$. – RRL Mar 22 '15 at 18:35
• 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. – RRL Mar 22 '15 at 19:18
• 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*} – Sujaan Kunalan Mar 22 '15 at 23:03
• @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. – 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 $$\int_E f=\sup\{\int_E\varphi:\varphi \text{ simple }, 0\leq \varphi\leq f\}=\infty.$$