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In the treatise of analysis volume 2 second edition of Dieudonné, there is an exercise which I found not easy because of a wrong & misleading hint. Can somebody tell me if I missed something ?

The exercise number is 13.9.25.a)

Here is the text of the 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 $p\geq n\geq 0$ :

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


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

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.

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I can't quite follow your logic in the equation after "By definition of a martingale, we get" and also after "Therefore, we deduce that". But you might like to know that this is a special case of a standard theorem in probability, called the martingale convergence theorem and due to Doob. You should be able to find a proof in any measure-theoretic probability textbook that deals with martingales. – Nate Eldredge Feb 19 '12 at 14:45
@NateEldredge Thank you for pointing me that it is a standard theorem ! About my proof : when I say "by definition of martingale" it is because for $i\leq m$ we get – brunoh Feb 19 '12 at 16:02
@NateEldredge (following of my previous comment) by definition $\int_{\left\{x\in X\ ;\ f_{n}(x)<b, \cdots, f_{i-1}(x)<b,\ f_{i}(x)\geq b\right\}}f_id\mu=\int_{\left\{x\in X\ ;\ f_{n}(x)<b, \cdots, f_{i-1}(x)<b,\ f_{i}(x)\geq b\right\}}f_md\mu$ therefore the inequality. – brunoh Feb 19 '12 at 16:10
@NateEldredge Concerning the "we deduce that" it is because exactly the same reasoning used on $A_n^m$ applies to the smaller set $A_n^m\cap B_q^p\cap A_s^r$. Sorry for being not clear. – brunoh Feb 19 '12 at 16:15
@NateEldredge Even if it is special case of a more general theorem, I would very much like to know if my proof works well in this special case, and if the hint was actually wrong. Thank you in advance. – brunoh Feb 19 '12 at 16:17

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