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I rewrote the end of the argument, hoping it is more clear. If not, please tell me at which point you do not understand.

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edited Mar 28, 2014 at 10:25

Thierry de la Rue

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 $T^{n-1}x\in B$, hence such. 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 $a_{n+1}>M$. But this prevents $x$ from being in the

The sets you defineconsider in 1. and 2. are of the form (in${\cal I}(i_n, v_n, M)$ where the sequence $(i_n)$ never hits two consecutive integers. In these sets, we cannot see two successiveonly the numbers $a_n$ strictly larger that$a_{i_n}$ are allowed to exceed $M$), hence ${\cal I}(i_n, v_n, M)\cap C=\emptyset$. It follows that Then these sets have zero${\cal I}(i_n, v_n, M)$ are included in the complement of $C$ which is $\mu$-measurenegligible, henceand it follows that they havee zero Lebesgue measure.

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 $T^{n-1}x\in B$, hence such that both $a_n>M$ and $a_{n+1}>M$. But this prevents $x$ from being in the sets you define in 1. and 2. (in these sets, we cannot see two successive $a_n$ strictly larger that $M$). It follows that these sets have zero $\mu$-measure, hence zero Lebesgue measure.

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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created Mar 27, 2014 at 14:58

Thierry de la Rue

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 $T^{n-1}x\in B$, hence such that both $a_n>M$ and $a_{n+1}>M$. But this prevents $x$ from being in the sets you define in 1. and 2. (in these sets, we cannot see two successive $a_n$ strictly larger that $M$). It follows that these sets have zero $\mu$-measure, hence zero Lebesgue measure.