This is the theorem: 
Let $X$ be separable metric space endowed with non-atomic Borel measure such that $\mu X = 1$.
Using this theorem We can establish isomorphism between $X$ and $[0;1]$.
Denote this mapping by $f$ .
I want to show that for at least one positive-measured interval $I \subset [0;1], \: \:$$f^{-1}(I\setminus I’) \subset B$, where $I’$ is the subset of $I$, of which measure is equal to $0$, i.e. $m(I’) = 0$,
where $B$ is any ball in $X$ with positive measure.
To sum it up I have to show that Inside every positive-measured $B \subset X$, We would have the preimage of at least one subset of positive-measured interval taken out from $[0;1]$, of which Lebesgue measure would be equal to the measure of taken interval.
Is it possible to construct such isomorphic mapping or at least a measure-preserving mapping which would satisfy these conditions? Any help would be appreciated.
Let $X=[0,1]$ endowed with the Lebesgue measure. Define the bijection $f$ of $X$ such that for each $x\in X$ we have $f(x)=x$, if $x$ is irrational, and $f(x)=1-x$, if $x$ is rational. It is easy to check that the map $f$ preserves measure. Let $B$ be the ball of radius $1/2$ centered at $0$. Then $B=[0,1/2]$, $B$ has positive measure, but there is no interval $I\subset [0,1]$ of positive measure such that $f^{-1}(I)\subset B$.