Lebesgue measure of some set of irrational numbers

Question

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 $\mathcal{I}(n^{3}, n^{2}, M)$?

Answer 1

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.