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



Lebesgue measure of some set of irrational numbers
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.


