Can a continuous surjection from a Hilbert cube to a segment behave bad wrt Lebesgue measures? Suppose $\hat{I}$ is a Hilbert cube and $I$ is a line segment. 
Consider Lebesgue measures $m_1$ and $m_2$ on $\hat{I}$ and $I$ correspondingly. 
Let $f:\hat{I}\to I$ be a continuous surjection. 


*

*Is it true that if $m_1(K) = 1$ for a Borel set $K$ then
$m_2(f(K) ) = 1$ ? 

*If the answer to the first question is no, would
    it help if I assume in addition that $f$ is Hölder continuous
    (with respect to metrics that are compatible with Borel structures)?


This is a continuation of this question.
P.S. This is the third question in the series, I promise this one is the last.
 A: I got an answer from MathOverflow. A counterexample can be constructed using  The Devil's staircase. If one replaces $\hat{I}$ by $I$, than the Cantor function is Holder continuous surjection and it is easy to construct a full measure set K (the set where the function is locally constant = the union of middle third intervals = the set of numbers whose base 3 expansion contains at least one 1) that is mapped to a zero measure set (the set of numbers with base 2 expansion having only finite nonzero coordinates). 
To understand that $K$ has measure $1$ look at $I\setminus K$. It is a set of numbers without a single $1$ in base-3 expansion. I will show that this set has measure zero due to standard ergodic theory.  Base $3$ expansion of a number $y$ can be contructed by assigning a sequence of digits $\{\sigma_k\}$ from $\{0,1,2\}^\mathbb{N}$ to the trajectory of $y$ under the action of $E_3 (x) = 3x (\mod 1)$. 
If $E_3^{k} (x) < 1/3$, put $\sigma_k = 0$, if  $1/3 < E_3^{k} (x) < 2/3$, put $\sigma_k = 1$, and so on.
As $E_k$ is ergodic with respect to Lebesgue measure for every $k\geq 2$, for Lebesgue almost every $x$ the trajectory of $x$ under the action of $x \to 3x  (\mod 1)$ visits every open set (including middle third).
To understand that $f(K)$ has zero measure one can use a similar argument -- for Lebesgue almost every $x$ the trajectory of $x$ under the action of $x \to 2x  (\mod 1)$ visits every open set (including second half) infinitely many times.
To handle the case of Hilbert cube one can take a composition of the Cantor function and a projection.
In fact, there are other similar examples here
