Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I was looking again at the beautiful and quite complete work of Dieudonné, his Treatise of Analysis, to refresh my memory about some aspects of classical analysis. I especially love this Treatise for the quality of its exercises. Unfortunately, some of them are marred with plain mistakes or wrong hints, and make me waste a lot of time detecting all of them. Does anybody know a reliable errata for the volume 2 of this treatise ?

Here below are two instances of plain wrong hints in an exercise, and a sketch of a solution I found for each of them.

I would very much appreciate if somebody could give some kind of reassurance that I am on the right track here and that my solutions are correct.

Here is the text of the first exercise: enter image description here

The hint is obviously wrong because it is not possible to have the inequality $b\mu(A\cap F_q)\leq a\mu(A)$ for $q=n$ for instance.

Below is a solution I found for comments

Notation for $p\geq n\geq 0$ :

$$\begin{align*} A_{n}^p &=\left\{x\in X\ ;\ \sup_{p\geq r\geq n}f_{r}(x)\geq b\right\} \\ B_{n}^p &=\left\{x\in X\ ;\ \inf_{p\geq r\geq n}f_{r}(x)\leq a\right\} \\ \end{align*}$$

and

$$\begin{alignat*}{2} A_{n} &=\bigcup_{p\geq n} A_{n}^p &\quad B_{n} &=\bigcup_{p\geq n} B_{n}^p \\ A &=\bigcap_{n\geq 0}A_{n} &\quad B &=\bigcap_{n\geq 0}B_{n} \end{alignat*}$$

Then we have $E_{ab}=A\cap B$. We also note that the unions and intersections in the previous definitions are respectively increasing and decreasing sequences.

Choose $r\geq s\geq p\geq q\geq m\geq n$. First we notice that $$A_{n}^m = \bigcup_{m\geq i\geq n} \left\{x\in X\ ;\ f_{n}(x)<b, \cdots, f_{i-1}(x)<b,\ f_{i}(x)\geq b\right\}$$ the union being of disjoints sets. By definition of a martingale, for $i\leq m$, we have $$\int_{x\in X\ ;\ f_n(x)<\cdots,f_{i−1}(x)<b,\ f_i(x\geq b} f_i\ d\mu=\int_{x\in X\ ;\ f_n(x)<\cdots,f_{i−1}(x)<b,\ f_i(x\geq b}f_m\ d\mu$$ Then, we get $$\int_{A_n^m}f_{m}d\mu \geq b\mu(A_{n}^m)$$ For the same reasons, we also get $$\int_{B_{n}^m}f_{m}d\mu \leq a\mu(B_{n}^m)$$

Therefore, we deduce that $$\int_{A_{n}^m\cap B_{q}^p \cap A_{s}^r}f_{r} d\mu \geq b\mu(A_{n}^m\cap B_{q}^p \cap A_{s}^r)$$ and $$\int_{A_{n}^m\cap B_{q}^p \cap A_{s}^r}f_{r}d\mu \leq \int_{A_{n}^m\cap B_{q}^p}f_{r}d\mu =\int_{A_{n}^m\cap B_{q}^p}f_{p}d\mu \leq a\mu(A_{n}^m\cap B_{q}^p) \leq a\mu(A_{n}^m\cap B_{q})$$

Let $$b\mu(A_{n}^m\cap B_{q}^p \cap A_{s}^r) \leq a\mu(A_{n}^m\cap B_{q})$$

By successively having $r$ then $s$ then $p$ then $q$ goes towards infinity we get $$b\mu(A_{n}^m\cap B \cap A)\leq a\mu(A_{n}^m\cap B)$$ If $m$ then $n$ goes towards infinity we get $$b\mu(E_{ab})\leq a\mu(E_{ab})$$ QEA.

Here is the text of the second example of wrong hint in an exercise

enter image description here

and a solution I found

Firstly, here is a counterexample for the impossibility to build a sequence $(B_{n})$ as suggested in the book: For $k\geq 1$ et $0\leq n< 2^k$, let $J_{n}^k=[n2^{-k},(n+1)2^{-k}]$. We take for $n\geq 0$ : $$A_{n}=\bigcup_{j\in 2N, 0\leq j<2^{n+1}}J_{j}^{n+1}$$ Then $n\geq 0$, $\lambda(A_{n})=\frac{1}{2}$ et $I_{n}=I$. We rightly have $r\geq n$ and $J_{j}^n \subset I_{n}$, $\lambda(A_{r}\cap J_{j}^n)=\frac{1}{2}\lambda(J_{j}^n)$.

But for all subsequences $(A_{n_{k}})$, we get $$\bigcap_{k\geq0}A_{n_{k}}\cap J_{1}^{n_{0}+1}\subset A_{n_{0}}\cap J_{1}^{n_{0}+1}=\emptyset$$

Proof of the main result

For the proof of the main result, we set $D=\cap_{k\geq 1}I_{k}$, and we notice that $\lambda(D)\geq \frac{1}{2}m$ and for $r\geq n\geq 0$, we get $\lambda(A_{r}\cap I_{n})\geq \frac{1}{4}m^2$.

We then use the following lemma

If $C\subset D$ is a compact of measure $r>0$, then for all $0<\alpha<1$ there exist $k>0$ and a $J_{n}^k \subset D_k$ such as $$\lambda(C\cap J_{n}^k) > \alpha \lambda(J_{n}^k) \quad (*)$$ where $D_k$ here corresponds to the $D_k$ in text of the exercise.

Proof of the lemma

Cover $C$ by $N$ non-empty open disjoint intervals $(U_{i})$ with $1\leq i\leq n$ such that $$r\leq \sum_{i}\lambda(U_{i}) < r+\epsilon$$ For $k$ big enough (for example such as $2^{-k}$ is smaller than half the smallest length of the intervals $(U_{i})$), let $$V_{k}=\{0\leq n< 2^k ; \exists 1\leq i\leq N \quad J_{n}^k\subset U_{i}\}$$ Then we get $$0\leq \sum_{1\leq i\leq N}\lambda(U_{i}) -\sum_{n\in V_{k}} \lambda(J_{n}^k) < 2^{-k+1}N$$ and $$0\leq \sum_{n\in V_{k}} \lambda(J_{n}^k) -\sum_{n\in V_{k}} \lambda(J_{n}^k\cap C)< \epsilon$$ If the lemma were false, then we have $$r\leq \sum_{1\leq i\leq N}\lambda(U_{i}) < 2^{-k+1}N+\epsilon+\alpha\sum_{n\in V_{k}} \lambda(J_{n}^k)\leq 2^{-k+1}N+\epsilon+\alpha\sum_{1\leq i\leq N}\lambda(U_{i})$$ soit $$(1-\alpha)r\leq 2^{-k+1}N+(1+\alpha)\epsilon$$ which is absurd as soon as $k$ is big enough and $\epsilon$ small enough.

This lemma has the following consequences

  • we always have $J_{n}^k\subset I_{k}$
  • as soon as $k'>k$, we can choose a $J_{n}^{k'}$ in $I_{k'}$ which fulfills $(*)$ and such as $J_{n}^{k'}\subset J_{n}^k$
  • with $\alpha$ close enough to $1$ we get as soon as $r\geq k$ $$\lambda(C\cap A_{r}\cap J_{n}^k)>0$$
  • if $k$ is big enough, using again the lemma for $C\cap (I-J_{n}^k)$, we see we can get the previous inequality in two disjoint closed intervals after a certain rank $k'>k$.

The proof is then not too difficult to finish by a recurrence building disjoint sequences of decreasing intervals of the type $J_{n}^k$, for which intersections are the points of $I$, but I lack space to write it in detail !

share|improve this question
add comment

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.