If I understood the question correctly, then I assume you require $\mu$ to be such that $\mu(E)\geq x_{i}$ for all those $i\in \mathbb{N}$ for which $y_{i}\in E$.
In particular we will want $m(f^{-1}(\{y_{i}\}))\geq x_{i}$ for every $i\in \mathbb{N}$. We use the fact that $\sum_{i=1}^{\infty}x_{i}=1$ and we choose $x_{0}:=0$. Let $f:[0,1]\to \mathbb{R}$ be such that:
$[0,x_{1})\mapsto y_{1}$
$[x_{1},x_{1}+x_{2})\mapsto y_{2}$
$[x_{1}+x_{2},x_{1}+x_{2}+x_{3})\mapsto y_{3}$
... etc (And let $f(1):=z$ for some $z\in\mathbb{R}\setminus\{y_{i}:i\in\mathbb{N}\}$)
In other words, we set $f(x)=y_{i}$ if $x\in [\sum_{n=0}^{i-1}x_{n},\sum_{n=0}^{i}x_{n})$. Hence the preimage of each $y_{i}$ is the interval $[\sum_{n=0}^{i-1}x_{n},\sum_{n=0}^{i}x_{n})$ which has Lebesgue measure of $x_{i}$, which guarantees that the condition that you needed holds. Hence $\mu(E)=\sum_{y_{i}\in E}x_{i}\geq x_{i}$ for those $i$ when $y_{i}\in E$.
The preimage of each open subset $U\subset \mathbb{R}$ is a Borel set of $[0,1]$, so $f$ is a measurable function.