I aim to show that $\int_{(0,1]} 1/x = \infty$. My original idea was to find a sequence of simple functions $\{ \phi_n \}$ s.t $\lim\limits_{n \rightarrow \infty}\int \phi_n = \infty$. Here is a failed attempt at finding such a sequence of $\phi_n$:

(1) Let $A_k = \{x \in (0,1] : 1/x \ge k \}$ for $k \in \mathbb{N}$.

(2) Let $\phi_n = n \cdot \chi_{A_n}$

(3) $\int \phi_n = n \cdot m(A_n) = n \cdot 1/n = 1$

Any advice from here on this approach or another?

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    $\begingroup$ You could use that the integrand is continuous and positive on the interval, so coincides (and co-exists, so exists iff exists) with the (improper) Riemann integral $\int_0^1 \frac{\mathrm dx}x$. $\endgroup$ – Lord_Farin Oct 31 '12 at 15:29
  • $\begingroup$ You are forgetting the relationship between the sequence $\phi_n$ and the function $x \mapsto 1/x$. $\endgroup$ – Mercy King Oct 31 '12 at 15:33
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    $\begingroup$ If that Lebesgue integral exists, it is greater than all the integrals $\int _{1/n}^1 1/x dx$ by positivity· $\endgroup$ – Matthew Towers Oct 31 '12 at 15:37
  • $\begingroup$ I changed \underset{n\to\infty}{lim} to \lim\limits_{n\to\infty}. The first looks like this: $\underset{n\to\infty}{lim}$. The second looks like this: $\lim\limits_{n\to\infty}$. The difference is not only that $\lim$ is not italicized, but also the preceding and following spacing in things like $a\lim b$. Also, when it is in a "displayed" setting rather than an "inline" setting, the subscript will appear directly under $\lim$ without the use of \limits. \limits is also used with things like \sum to change things like $\sum_{i=1}^n$ to $\sum\limits_{i=1}^n$. $\endgroup$ – Michael Hardy Oct 31 '12 at 15:46
  • $\begingroup$ @MichaelHardy I still prefer the command \displaystyle to enforce correct placements of sub- and superscripts. $\endgroup$ – Lord_Farin Oct 31 '12 at 15:54

Write $I_k:=((k+1)^{-1},k^{—1})$. Then for each $n$, $s_n:=\sum_{k=1}^nk\chi_{I_k}$ is a simple non-negative function, and $0\leq s_n\leq f(x):=1/x$. We have $$\int_{(0,1]}s_n \, d\lambda=\sum_{k=1}^nk\left(\frac 1k-\frac 1{k+1}\right)=\sum_{k=1}^nk\frac{k+1-k}{k(k+1)}=\sum_{k=1}^n\frac 1{k+1}.$$ So $$\int_{(0,1]}s_{2n} \, d\lambda-\int_{(0,1]}s_n \, d\lambda=\sum_{k=n+1}^{2n}\frac 1{k+1}\geq\frac n{2n+1}\geq \frac 13.$$ As the sequence $\{\int_{(0,1]}s_n \, d\lambda\}$ is increasing, it has a limit. This one can't be finite by the last inequality, and the sequence is non-negative, so it converges to $+\infty$. This proves that

$$\sup\{\int_{(0,1]}s \, d\lambda,0\leq s\leq f, s\text{ simple}\}$$

is infinite, that is, $f$ is not Lebesgue integrable.

  • $\begingroup$ Can you give me the link to the associated theorem using which you have proved that f is not Lebesgue integrable? Thanks! $\endgroup$ – Sriram Natarajan Aug 3 '16 at 9:15
  • $\begingroup$ @SriramNatarajan It is just the definition of Lebesgue integral of a non-negative measurable function. $\endgroup$ – Davide Giraudo Aug 3 '16 at 9:25
  • $\begingroup$ @DavideGiraudo, can we apply this method to any $1/x^{\alpha}$, with $\alpha\ge 1$? $\endgroup$ – roi_saumon Nov 11 '20 at 21:56
  • $\begingroup$ @DavideGiraudo, or just use $1/x^{\alpha}\ge 1/x$ on $(0,1]$ to conclude for those cases? $\endgroup$ – roi_saumon Nov 11 '20 at 21:58

I think this may be the same as what Davide Giraudo wrote, but this way of saying it seems simpler. Let $\lfloor w\rfloor$ be the greatest integer less than or equal to $w$. Then the function $$x\mapsto \begin{cases} \lfloor 1/x\rfloor & \text{if } \lfloor 1/x\rfloor\le n \\[8pt] n & \text{otherwise} \end{cases}$$ is simple. It is $\le 1/x$ and its integral over $(0,1]$ approaches $\infty$ as $n\to\infty$.

  • $\begingroup$ how can one show that the integral of the simple function approaches infinity? $\endgroup$ – mathjacks Mar 25 '15 at 2:21
  • $\begingroup$ @mathjacks : That integral is a finite sum: $\left(\dfrac 1 2 + \dfrac 2 6 + \dfrac 3 {12} + \dfrac 4 {20} + \cdots + \dfrac n {n(n+1)}\right) + \dfrac n {n+1} $. One gets, for example, $\dfrac 4 {20}$ from the fact that the value is $4$ between $1/5$ and $1/4$, and the length of that interval is $\dfrac 1 4 - \dfrac 1 5 = \dfrac 1 {20}$. So we're talking about divergence of the harmonic series. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 17 '15 at 22:59

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