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Let $(i_{n})$ be a strictly increasing sequence of natural numbers, $(v_{n})$ be an unbounded sequences of natural numbers and $M\geq 2$. Denote by $\mathcal{I}(i_{n}, v_{n}, M)$ the set of all irrational numbers $\alpha=[a_{1}, a_{2},...,a_{s},...)\in (0,1)$ which is $a_{i_{n}}\leq v_{n}$ and $a_{s}\leq M$ for any $s\in \mathbb{N}\setminus \{i_{n}, n=1,2,...\}.$

My question related to the Lebesgue measure of this set. Of course this measure depends on exact form of sequences $i_{n}$ and $v_{n}.$ For example in my cases:

  1. What is the Lebesgue measure of $\mathcal{I}(n^{4}, n^{6}, M)$?

  2. What is the Lebesgue measure of $\mathcal{I}(n^{3}, n^{2}, M)$?

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  • $\begingroup$ I don't understand how you are defining $\alpha$. $\endgroup$ – Ben Willson Mar 27 '14 at 2:11
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    $\begingroup$ What does the statement of Question 1 mean? Maybe you want to say $\mathcal{I}(n^2,n,M)$? In any case, you may want to reconsider the answers to your question from yesterday, they might give you some advice on today's problem as well. $\endgroup$ – Kurisuto Asutora Mar 27 '14 at 2:17
  • $\begingroup$ I'm fairly sure that fixing a partial quotient or insisting that it be less than a fixed bound will in each case reduce the measure by a factor bounded away from unity. You have an infinite number of such constraints so it is clear the answer should be zero. Perhaps an expert can confirm and provide more details. $\endgroup$ – user25199 Mar 27 '14 at 7:22
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    $\begingroup$ @user39115 Look up "Continued fraction expansion." $\endgroup$ – user25199 Mar 27 '14 at 11:54
  • $\begingroup$ It is not clear to me what does it mean $\alpha=[a_1,a_2,\ldots,a_s,\ldots)$? If the meaning is $\alpha=\sum_{i=1}^{\infty} \frac{a_i}{10^i},$ then the condition $a_s<M$ does not make sense for $M>9,$ so probably there will be not lost of generality in assuming $M=2$ for example, in which case i will bet that is not difficult to prove that the answer is a number in (0,1). If the meaning is the continuous fraction expansion then I will bet again that is not difficult to prove that the answer is $0.$ $\endgroup$ – user39115 Mar 27 '14 at 11:58
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The continued fraction expansion is related to the Gauss transformation $T:(0,1)\to(0,1)$, defined by $$ Tx:=\frac{1}{x} \mod 1. $$ (Indeed, if $x=[a_1,a_2,\ldots)$, then $Tx=[a_2,a_3,\ldots)$.)

It is well known that $T$ admits an absolutely continuous invariant probability measure $\mu$, given by $$ \mu(A):=\frac{1}{\ln 2}\int_A \frac{dx}{1+x}, $$ and that $T$ is ergodic for $\mu$.

Now, given $M\ge 2$, the set $B$ of $x\in(0,1)$ for which both $a_1$ and $a_2$ are stricly larger than $M$ clearly satisfies $\mu(B)>0$. Hence, by ergodicity, for $\mu$-almost every $x$ there exist infinitely many integers $n$ such that $T^{n-1}x\in B$. This exactly means that $\mu(C)=1$, where $C$ is the set of numbers $x\in(0,1)$ for which there exist infinitely many integers $n$ satisfying both $a_n>M$ and $a_{n+1}>M$.

The sets you consider in 1. and 2. are of the form ${\cal I}(i_n, v_n, M)$ where the sequence $(i_n)$ never hits two consecutive integers. In these sets, only the numbers $a_{i_n}$ are allowed to exceed $M$, hence ${\cal I}(i_n, v_n, M)\cap C=\emptyset$. Then these sets ${\cal I}(i_n, v_n, M)$ are included in the complement of $C$ which is $\mu$-negligible, and it follows that they havee zero Lebesgue measure.

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  • $\begingroup$ Thank you very much. Sorry, the conclusion not so clear for me. Could you explain me a bit more in detail? $\endgroup$ – sokho Mar 28 '14 at 4:08

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