Let $f$ be a real-valued monotone-increasing function, $\kappa$ a sequential covering class of $\mathbb{R}$ consisting of the empty set $\emptyset$ and the open intervals $I_{a,b}=(a,b)$, where $a$ and $b$ are real valued, and we define the set function:
$$ \lambda(I_{a,b}) = f(b) - f(a), $$ then the Lebesgue-Stiltjes outer measure induced by $f$ is given by:
$$ \mu^*_f(A) = \inf\{\sum_{n=1}^\infty \lambda(E_n): E_n \in \kappa, A \subset \cup_{n=1}^\infty E_n\}, $$
where $A$ is an arbitrary subset of $\mathbb{R}$. If $f(x)=x$ we get the Lebesgue outer measure. My question is if a set is Lebesgue-Stiltjes (outer) measurable if and only if it is Lebesgue measurable? I considered the definition of measurable sets, where:
$$ \mu^*_f(A) \geq \mu^*_f(A\cap E) + \mu^*_f(A -E), $$
implies that subset $E$ of $\mathbb{R}$ is measurable given that $A$ is any subset of $\mathbb{R}$. I tried assuming e.g. that $f(x) \geq x$ i which case we would have that
$\mu^*_f(A) \geq \mu^*(A) \geq \mu^*(A\cap E) + \mu^*(A -E)$
but got stuck and couldn't come up with a good argument, especially not for the general case.