If consider the map $f:2^{\omega}\to [0,1]$ given $f(x)=\sum_{i=0}^{n}x(i)2^{-i-1}$. Let $\nu_{}$ be the Haar measurable on $2^{\omega}$. Then $f\nu_{}=\big.m\big|_{[0,1]}$ where $m$ is the Lebesgue measure.
Can give me any hint for prove this?, thank you very much to give me a suggestion.
Note that the diadic interval $A=\left[\frac{k}{2^n}, \frac{k+1}{2^n} \right]$ is the $f$-image of the set $$B= \left\{ (a_i)\in 2^\omega\ \left| \ \ \sum_{i=0}^{n-1} a_i 2^{n-1-i} = k \right. \right\}.$$ Therefore $f\nu(A) = \nu(B)= 1/2^n$, which agrees with Lebesgue measure. I claim that the diadic intervals generate the $\sigma$-algebra, and so $f\nu$ agrees with Lebesgue measure on a generating set, which establishes the result.