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.

This is not homework. Problem 3-38 reads:

Let $A_{n}$ be a closed set contained in $(n,n+1)$. Suppose that $f:\mathbb{R}\rightarrow \mathbb{R}$ satisfies $\int_{A_{n}}f=(-1)^{n}/n$ and $f(x)=0$ for $x\notin$ any $A_{n}$. Find two partitions of unity $\Phi$ and $\Psi$ such that $\sum_{\phi\in\Phi}\int_{\mathbb{R}}\phi\cdot f$ and $\sum_{\psi\in\Psi}\int_{\mathbb{R}}\psi\cdot f$ converge absolutely to different values.

A few observations: First, $n\ge 1$. Second, Spivak uses what he calls an extended integral, whose definition and relations with the usual integral can be found on p.65, which can be found here or here.

It may be helpful to have an example of such a function in mind. Let $A_{n}=$ closed interval of length $1/2n$ centered at the point $(2n+1)/2$. Clearly $A_{n}\subset(n,n+1)$. then define $$f(x)=\begin{cases} \hphantom{-}2& \text{if $x\in A_{n}$ for $n$ even}\\ -2& \text{if $x\in A_{n}$ for $n$ odd}\\ \hphantom{-}0& \text{otherwise}. \end{cases}$$

A possible approach: Let $a_{n}=(-1)^{n}/n$. Since $\sum_{n}a_{n}=\alpha\in\mathbb{R}$ but the convergence is conditional, then for any $\beta\not=\alpha$ there is a rearrangement $\{b_{n}\}$ of the sequence $\{a_{n}\}$ such that $\sum_{n}b_{n}=\beta$.

Now, we form a family of open sets $\{U_{n}\}$, where $U_{n}$ is the union of $n$ intervals $(k,k+1)$, each corresponding to a term of the $n$-th partial sum of $\sum_{n}a_{n}$. We form a similar family $\{V_{n}\}$ looking at the partial sums of $\sum_{n}b_{n}$. E.g., if we let $\{b_{n}\}=\{-1,1/2,1/4,-1/3,1/6,1/8,-1/5,\ldots\}$ we have $V_{3}=(1,2)\cup(2,3)\cup(4,5)$ while since $\{a_{n}\}=\{-1,1/2,-1/3,1/4,\ldots\}$ we have $U_{3}=(1,2)\cup(2,3)\cup(3,4)$.

If we slightly fatten-up the $U_{n}$ (resp. the $V_{n}$) we form open covers $\mathcal{U}$ (resp. $\mathcal{V}$) of all the reals greater or equal than 1 without ading points where $f$ in non-zero. My heart tells me that partitions of unity $\Phi$ and $\Psi$ subordinate to $\mathcal{U}$ and $\mathcal{V}$, respectively, will be the desired one. But alas I am lost!

Does any one know how to show that the aforementioned partitions of unity are the desired ones?

Other possible approaches to the solution are also welcomed.

In addition to two posts linked above, related issues with other problems and statements about integration in Spivak´s book can be found here and here.

share|improve this question

1 Answer 1

Since Spivak requires absolute convergence, it is not enough to rearrange the series. We need a "blocking" lemma of the following kind:

Lemma. Suppose $\sum a_n$ is a conditionally convergent series. Then for every $\beta\in \mathbb R$ there exists a partition $\mathbb N=\bigcup_{j=1}^\infty B_j$ where each set $B_j$ is finite, and the series $\sum_j \left(\sum_{n\in B_j} a_n\right)$ converges absolutely to $\beta$.

Assume the lemma for now. Our partition should include open sets $U_j=\bigcup_{n\in B_j}(j,j+1)$. These do not cover the integers, though. Cover each integer by a tiny interval $(n-\epsilon,n+\epsilon)$ where $\epsilon$ is small enough so that the interval does not meet the sets $A_n$ and $A_{n-1}$. Note that $f$ is identically zero within such intervals. For each $n$, the set $A_n$ meets exactly one element $U_j$ of our open cover; therefore, the corresponding function $\varphi_j$ from a partition of unity will be identically $1$ on $A_n$. We have $\int f\varphi_j=\sum_{n\in B_j} a_n$ which gives the desired result. $\quad\Box$

As for the lemma, the proof goes like this: first, rearrange the series so that it converges conditionally to $\beta$; then divide the rearranged series $\sum a_{\sigma(k)}$ into blocks $N_i\le \sigma(k)<N_{i+1}$, $i=1,2,3,\dots$ such that the sum within the $i$th block is at most $2^{-i}$ for all $i\ge 2$. (Idea: cut off the tail repeatedly when it gets smaller than $2^{-i}$.)

share|improve this answer

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.