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Problem 3-37 (b) reads:

Let $A_{n}=[1-1/2^{n},1-1/2^{n+1}]$. Suppose that $f:(0,1)\rightarrow \mathbb{R}$ satisfies $\int_{A_{n}}f=(-1)^{n}/n$ and $f(x)=0$ for any $x\notin$ any $A_{n}$. Show that $\int_{(0,1)}f$ does not exist, but ${\displaystyle\lim_{\varepsilon\to 0}}\int_{(\varepsilon,1-\varepsilon)}f=\log\,2$.

Note: First that $n\ge 1$ and second that Spivak probably forgot a minus sign, i.e. he intended us to show that ${\displaystyle\lim_{\varepsilon\to 0}}\int_{(\varepsilon,1-\varepsilon)}f=-\log\,2$.

Note: The integrals are to be understood as extended integrals (see p.65 of the book or the image below). Theorem 3-12 (3) links this notion of integral with the old one.

That $\int_{(0,1)}f$ does not exist is not hard to show.

However, it seems to me that ${\displaystyle\lim_{\varepsilon\to 0}}\int_{(\varepsilon,1-\varepsilon)}f=\log\,2$ is not necessarily true. We don´t know how $f$ behaves outside the sets $A_{n}$ and so it may very well be that $\int_{(\varepsilon,1-\varepsilon)}f$ does not exist.

Am I correct in guessing that we need additional hypotheses?

If we assume additional properties of $f$, e.g. that $f$ does not change sign within each set $A_{n}$ and is bounded on each set $A$ then Theorem 3-12 applies and the result is true.

Related issues with other problems and statements about integration in Spivak´s book can be found here and here. p.65 of the book.

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Um, am I wrong saying that $[0,1[ = \bigcup_{n=0}^{\infty} A_n$? Because then it doesn't make sense to say "for any $x \notin \bigcup_{n=0}^{\infty} A_n$" because it concerns no $x \in ]0,1[$. –  Patrick Da Silva Nov 25 '12 at 14:08
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I fail to see the problem. The statement, as you write it, says that $f=0$ outside the $A_n$. –  Martin Argerami Nov 25 '12 at 15:09
    
I stated the problem exactly as it is the book. I know there is no need for additional asumptions to show that the integral does not exist. Also given the fact that Spivak explicitly says "$f(x)=0$ for $x\not\in$ any $A_{n}$" I am inclined to think that he meant that $n\ge 1$. –  John Nov 25 '12 at 17:11
    
@John : The only way the $f(x) = 0$ for $x \notin A_n$ would make sense is indeed if $n \ge 1$. Otherwise I don't see the point. –  Patrick Da Silva Nov 25 '12 at 18:01

2 Answers 2

You don't need any additional hypothesis. The reason why the limit doesn't exist is because when you consider a Riemann partition and let the step go to zero, you can still choose your partition nodes inside $A_{2n}$ if you want the integral to go to infinity or inside $A_{2n+1}$ if you want the integral to go to minus infinity, loosely speaking, so there is no problem.

However, when you consider the limit when $\varepsilon \to 0$ of the integral over $(\varepsilon, 1- \varepsilon)$, you are not allowed such freedom ; you are forced to integrate over each and every of the $A_n$ and take them one at a time ; this forces the integral to be close to a partial sum of the series $(-1)^n/n$, thus the limit converges to the series, which is $\log(2)$. I spoke vaguely on purpose ; I leave the details up to you.

Maybe you consider this as an additional hypothesis, but it is actually not ; you can suppose that $\int_{[a,b]} f$ exists for every interval $[a,b] \subseteq A_n$. This is because Riemann-integrability is preserved in sub-intervals over which $f$ is integrable, so since you are given that $f$ is integrable over $A_n$, it is also integrable over its subintervals, thus allowing you to compute the integrals over the intervals $(\varepsilon,1-\varepsilon)$.

Hope that helps,

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The definition of integral used in the problem involves partitions of unity and absolute convergence of a certain series, and may not coincide with the usual one. See p.65 of the book. Is that the integral you have in mind? Because you don´t mention partitions of unity at all. –  John Nov 25 '12 at 18:38
    
@John : If you use a different integral than the Riemann integral and do not specify which integral you use, I cannot guess that. I didn't read the book to answer your question ; I just assumed you were using the Riemann integral. Is the integral you are using standard? Does it have a name? I could look it up. –  Patrick Da Silva Nov 25 '12 at 18:44
up vote 0 down vote accepted

I want to note that integral means extended integral as it is defined on p.65 of the book.

Part a) (I am not completely sure this will work, criticism welcomed!)

For $k\ge 1$ we let $$U_{k}=\left(1-\frac{1}{2^{n}-1}, 1-\frac{1}{2^{n+1}+1}\right).$$ Consider a partition of unity $\Phi$ subordinate to the cover $\{U_{k}\}_{k\ge 1}$. By adding the $\varphi\in\Phi$ with the same $U$ in condition $(4)$ of Theorem 3-11 we may assume that there is only one function $\varphi_{k}$ for each $U_{k}$. Then when considering the convergence of $$\sum_{k=1}^{\infty}\int_{(0,1)}\varphi_{k}|f|$$ we have that $\sum_{k=1}^{n-1}\int_{A_{k}}|f|\leq\sum_{k=1}^{n}\int_{(0,1)}\varphi_{k}|f|$. But $\int_{A_{k}}|f|\ge 1/n$. Therefore, the series above diverges and so $\int_{(0,1)}f$ does not exist.

Part b)

If we assume that the sign of $f$ does not change within each $A_{n}$ and that $f$ is bounded on each $A_{n}$ then Theorem 3-12 (3) applies and $\int_{(\varepsilon,1-\varepsilon)}f$ is just the old Riemann integral $\int_{\varepsilon}^{1-\varepsilon}f(x)dx$ and so the result is clear, for we will have that $$\lim_{\varepsilon\to 0}\int_{\varepsilon}^{1-\varepsilon}f(x)dx=\sum_{n=1}^{\infty}\int_{A_{n}}f(x)dx=\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}=-\log\, 2.$$

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