# Another exercise in Measure Theory

In the treatise of analysis volume 2 second edition of Dieudonné, there is an exercise (13.8.8) which I found not easy because of a wrong & misleading hint.

Can somebody tell me if I missed something ?

Text of the exercise:

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 $(A_{n_{k}})_{k\geq 0}$, 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 $J_{n}^k\in \mathcal{D}_{k}$ such as $$\lambda(C\cap J_{n}^k) > \alpha \lambda(J_{n}^k) \quad (*)$$

Proof of the lemma

Cover $C$ by $N$ non-empty open disjoint intervals $(U_{i})_{1\leq i\leq N}$ such that $$r\leq \sum_{1\leq i\leq N}\lambda(U_{i})<r+\epsilon$$ 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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You should really add the exercise statement! I for one cannot view the pages you linked to. Also, links may not work anymore some time in the future, so in such case this question won't help anybody else, if you don't make it self-contained. –  Sam Feb 23 '12 at 18:50
@Sam You are certainly right ! But I do not know how to copy it ... Should I retype it entirely ? –  brunoh Feb 23 '12 at 18:53
Maybe I'm missing something, but the exercise requires proving the existence of the decreasing sequence $(I_k)_k$; so in your counterexample this sequence has to be arbitrary. How come you claim that $J_{j}^n \subset I_{n}$? –  Martin Argerami Feb 23 '12 at 18:56
I added the image with the exercise to the question, but I don't have edit privileges so I cannot make it public. –  Martin Argerami Feb 23 '12 at 19:05
@brunoh: in the edit window, there is a button that looks like a little landscape, and it lets you add the image. You can either authorize my edit (I don't know how that works) or you can add the image yourself. –  Martin Argerami Feb 23 '12 at 19:16