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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?

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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 $\int_0^1 \frac{\mathrm dx}x$.2012-10-31
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    You are forgetting the relationship between the sequence $\phi_n$ and the function $x \mapsto 1/x$.2012-10-31
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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·2012-10-31
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    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 $\sum\limits_{i=1}^n$.2012-10-31
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    @MichaelHardy I still prefer the command `\displaystyle` to enforce correct placements of sub- and superscripts.2012-10-31
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    @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.2012-10-31
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    @MichaelHardy I know :) That tends to be why I like it more. It's good to be aware of both options.2012-10-31
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    @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?2016-08-03

2 Answers 2

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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.

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    Can you give me the link to the associated theorem using which you have proved that f is not Lebesgue integrable? Thanks!2016-08-03
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    @SriramNatarajan It is just the definition of Lebesgue integral of a non-negative measurable function.2016-08-03
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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 $n\to\infty$.

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    how can one show that the integral of the simple function approaches infinity?2015-03-25
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    @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{}$2015-12-17