Your set has measure zero by theorems of Khinchin. First a theorem of Khinchin shows that for almost all real numbers $x$ (i.e. outside a set of measure zero) one has
$$
\lim_{n\to \infty} \frac{\log q_n}{n}= C
$$
for a positive constant $C$. Here $p_n/q_n$ are the convergents of $x$. So almost surely, the denominators $q_n$ are exponential in $n$. In fact the constant $C$ here is known to be $\pi^2/(12 \log 2)$ -- the Levy or Khinchin-Levy constant.
Next another Theorem of Khinchin shows that for almost all $x$ one has infinitely many approximations $p/q$ with
$$
\Big| x-\frac{p}{q}\Big| \le \frac{1}{q^2\log q}.
$$
Here what is used is that $\sum_q 1/(q\log q)$ diverges.
Putting both results together (and since continued fractions give best rational approximations) we see that for almost all $x$ we have infinitely many $n$ such that
$$
\Big| x-\frac{p_n}{q_n} \Big | \le \frac{C}{n q_n^2},
$$
for some positive constant $C$. Now recall that for any $x$ we have $|x-p_n/q_n|$ is about size $1/(q_nq_{n+1})$ and $q_{n+1}$ is about size $a_{n+1}q_n$. Therefore it follows from the above inequality that for infinitely many $n$ one has $a_n \ge cn$ for some constant $c>0$.
Edit: I just got the chance to look at Khinchin's book Continued Fractions, and he discusses this question explicitly in Theorem 30 of that book (page 63 in the Dover edition). Let $\phi(n)$ be an arbitrary positive function ($n\in {\Bbb N}$). Then the inequality $a_n \ge \phi(n)$ is satisfied for infinitely many $n$ for almost all real numbers $x$, provided $\sum 1/\phi(n)$ diverges. On the other hand if $\sum 1/\phi(n)$ converges, then this inequality is satisfied for only finitely many $n$ (and almost all $x$).
The condition $\lambda>1$ is sufficient and, at least almost, necessary:
To clarify, the space of irrational numbers $(0,1)-\mathbb Q$ is homeomorphic to
$\omega^\omega$ under the map that sends $\frac{1}{a_1+\frac{1}{a_2+\cdots}}$ to a function $f$ satisfying
$f(n)=a_{n+1}-1$. (As is well known.)
This way Lebesgue measure on $(0,1)$ induces a "Gaussian" measure on $\omega^\omega$.
Under this measure $\mu$, I claim that the following set has measure one:
$$\{f:f(n)\text{ is eventually bounded by $n^t$ for $t>1$, but not for
$t=1$}\}$$
Proof:
With $\delta_n=\sum_{k\le n^t}\epsilon_{nk}$ and
$\Delta_n=\delta_n\log 2$ and $\log=\log_e$, we have
$$\log\prod_{n=1}^\infty \sum_{k\le
n^t}\frac{\log\left(1+\frac{1}{k(k+2)}\right)}{\log
2}+\epsilon_{nk}=$$
$$\log\prod_{n=1}^\infty \frac{ \log 2+(\log (n^t+1)-\log(n^t+2))}{\log 2}+\delta_n =$$
$$\sum_{n=1}^\infty \log\left[ \log 2+(\log (n^t+1)-\log(n^t+2)) +\Delta_n \right]-\log\log 2$$
Here $\epsilon_{nk}=\pm\frac{A}{k(k+1)}e^{-\lambda\sqrt{n-1}}$ for
certain constants $A,\lambda$.
By comparison of the negative of this series with $\sum
\frac{1}{n^t}$, where we get a constant limit, the series
converges iff $t>1$:
$$\lim_{n\to\infty}\frac{\log\log 2-\log(\log 2+(\log (n^t+1)-\log
(n^t+2)+\Delta_n))}{n^{-t}}=$$
$$\lim_{n\to\infty}
\frac{(\log 2+(\log (n^t+1)-\log
(n^t+2)+\Delta_n))^{-1}[\frac{tn^{t-1}}{n^t+1}-\frac{tn^{t-1}}{n^t+2}+\Delta'(n)]}
{-tn^{-t-1}}=$$
$$\lim_{n\to\infty}\frac{-n^{t+1}[\frac{tn^{t-1}}{n^t+1}-\frac{tn^{t-1}}{n^t+2}+\Delta'(n)]}{t(\log 2+(\log (n^t+1)-\log
(n^t+2)+\Delta_n))}=$$
$$\lim_{n\to\infty}\frac{-n^{t+1}[\frac{n^{t-1}}{n^t+1}-\frac{n^{t-1}}{n^t+2}+\Delta'(n)/t]}{\log 2+(\log (n^t+1)-\log
(n^t+2)+\Delta_n)}=$$
$$\lim_{n\to\infty}\frac{-n^{2t}[\frac{1}{n^t+1}-\frac{1}{n^t+2}+\frac{\Delta'(n)}{tn^{t-1}}]}{\log 2+(\log (n^t+1)-\log
(n^t+2)+\Delta_n)}=$$
$$\lim_{n\to\infty}\frac{-n^{2t}[\frac{1}{(n^t+1)(n^t+2)}+\frac{\Delta'(n)}{tn^{t-1}}]}{\log 2+(\log (n^t+1)-\log
(n^t+2)+\Delta_n)}=$$
$$\frac{1}{\log 2}$$
provided $n^{t+1}\Delta'(n)\to 0$.
And indeed
$$n^{t+1}\Delta'(n)=(\pm A)(\log 2)n^{t+1}\frac{d}{dn}[e^{-\lambda\sqrt{n-1}}\sum_{k\le n^t}\frac{1}{k(k+1)}]$$
Ignoring the constant and using $f(n)=\sum_{k\le n^t}g(k)\approx
\int_1^{n^t}g(k)dk$, so $f'(n)\approx g(n^t)-g(1)$, we have
$$ n^{t+1} \left[ e^{-\lambda\sqrt{n-1}}[-\lambda\frac{1}{2\sqrt{n-1}}]f(n)+e^{-\lambda\sqrt{n-1}}f'(n) \right]= $$
$$ e^{-\lambda\sqrt{n-1}} n^{t+1} \left[ \left(-\lambda\frac{1}{2\sqrt{n-1}}\right)f(n)+f'(n) \right] $$
and this clearly goes to 0 as $n\to\infty$.
So $\log\prod=-\infty$ and hence $\prod=0$. This completes the proof.
Best Answer
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.