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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?

Michael Hardy
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user1770201
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    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 $\int_0^1 \frac{\mathrm dx}x$. – Lord_Farin Oct 31 '12 at 15:29
  • You are forgetting the relationship between the sequence $\phi_n$ and the function $x \mapsto 1/x$. – HorizonsMaths Oct 31 '12 at 15:33
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    If that Lebesgue integral exists, it is greater than all the integrals $\int _{1/n}^1 1/x dx$ by positivity· – Matthew Towers Oct 31 '12 at 15:37
  • 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 $\sum\limits_{i=1}^n$. – Michael Hardy Oct 31 '12 at 15:46
  • @MichaelHardy I still prefer the command \displaystyle to enforce correct placements of sub- and superscripts. – Lord_Farin Oct 31 '12 at 15:54
  • @Lord_Farin : One should remember that \displaystyle has more effects than just that. In this case, it will also alter the appearance of the integral sign. Maybe sometimes that's appropriate and sometimes not, but one should bear in mind that the effects are not identical. – Michael Hardy Oct 31 '12 at 15:57
  • @MichaelHardy I know :) That tends to be why I like it more. It's good to be aware of both options. – Lord_Farin Oct 31 '12 at 17:20
  • @Lord_Farin I can construct a series of increasing Reimann integrable functions converging to given function with limit of integrand tending to infinity. Why does that make the given function not a Lebesgue integrable one? – Sriram Natarajan Aug 03 '16 at 09:14

2 Answers2

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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.

Michael Hardy
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Davide Giraudo
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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 $n\to\infty$.

Michael Hardy
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  • how can one show that the integral of the simple function approaches infinity? – mathjacks Mar 25 '15 at 02:21
  • @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. ${}\qquad{}$ – Michael Hardy Dec 17 '15 at 22:59