Can a continuous surjection from a Hilbert cube to a segment behave bad wrt Lebesgue measures? Suppose $\hat{I}=[0,1]^\mathbb{N}$ is a Hilbert cube and $I=[0,1]$. 
Consider Lebesgue measures $m_1$ and $m_2$ on $\hat{I}$ and $I$ correspondingly. By Lebesgue measure on the Hilbert cube I mean the product
of Lebesgue measures on the segments.
Let $f:\hat{I}\to I$ be a continuous surjection. 
Let $K$ be a Borel subset of $\hat{I}$ with $m_1(K) = 1$.


*

*Is it true that $m_2(f(K) ) = 1$ ? 

*If the answer to the first question is no, what could be said if $f$ is a homeomorphism?

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


UPD:
Question 2 should be read: "... if $f$ is an open map?"
 A: I would like to comment on Joel David's answer and remark that his example also provides negative answers to questions 2 and 3 (even though for question 2, I can't see what you mean by "homeomorphism", since the two spaces are not homeomorphic).
You can observe that if $d:I\rightarrow I$ is the Devil's staircase, then $\frac12 id +\frac12 d$ is a non absolutely continuous homeomorphism of the interval, which is also Hölder continuous.
A: The answer to question 1 is no, not necessarily. The Devil's staircase is a continuous surjection $d:I\to I$, which has a measure one set $K_0\subset I$ with $m_2(d(K))=0$. The set $K_0$ is simply the open set of points on which $d$ is locally constant (the union of the middle third intervals). 

We may extend this example to a continuous surjection $f:\hat I\to I$ by defining $f(s)=d(s_0)$. The set $K=\{s\in \hat I\mid s_0\in K_0\}$ is determined by its first coordinate and has $m_1(K)=1$, but still $f(K)=d(K_0)$ having measure $0$. So it is a counterexample to question 1.
There is a version of the Devil's staircase, where one gives the flat regions a small but positive slope. The resulting map is  a modified staircase map $d^*:I\to I$, which is now a homeomorphism, with an open set $K_0\subset I$ of measure $1$, but where $d^*(K)$ does not have measure $1$. The set $K_0$ is again the middle third intervals. The corresponding function 
$f^*:\hat I\to I$ will now be a counterexample to statement 2.
