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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 !

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