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 (see for instance this question Separability of the set of positive measures) 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 one instance of a plain wrong hint in an exercise, and a sketch of a solution I found for it.
I would very much appreciate if somebody could give some kind of reassurance that I am on the right track here and that my solution is correct.
Here is the text of the exercise:
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.
