Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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 $x \mapsto 1/x$. –  Mercy Oct 31 '12 at 15:33
1  
If that Lebesgue integral exists, it is greater than all the integrals $\int _{1/n}^1 1/x dx$ by positivity· –  mt_ 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
show 2 more comments

2 Answers 2

up vote 4 down vote accepted

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 $$\int_{(0,1]}s_nd\lambda=\sum_{k=1}^nk\left(\frac 1k-\frac 1{k+1}\right)=\sum_{k=1}^nk\frac{k+1-k}{k(k+1)}=\sum_{k=1}^n\frac 1{k+1}.$$ So $$\int_{(0,1]}s_{2n}d\lambda-\int_{(0,1]}s_nd\lambda=\sum_{k=n+1}^{2n}\frac 1{k+1}\geq\frac n{2n+1}\geq \frac 13.$$ As the sequence $\{\int_{(0,1]}s_nd\lambda\}$ 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 $$\sup\{\int_{(0,1]}sd\lambda,0\leq s\leq f, s\mbox{ simple}\}$$ is infinite, that is, $f$ is not Lebesgue integrable.

share|improve this answer
add comment

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 $n\to\infty$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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