Lebesgue Integral on Lebesgue measurable set satisfies Caratheodory condition

Question

Let $f$ be a non-negative, measurable, and integrable over every compact set in $\Omega$, where $\Omega$ is an open set $\subset \mathbb{R}^d$.

For every Lebesgue measurable set $E$ (abbreviated as $\mu$-measurable from now on), define $$\nu(E) = \int_{E} f d\mu$$ where $\mu$ is Lebesgue measure, and $\int d\mu$ is Lebesgue integral.

For $E$ that is not $\mu$-measurable, define $$\nu(E) = \inf_{\mu-\text{measurable } F \supset E } \nu(F) \quad (\$)$$

Show that if $E$ is $\mu$-measurable in Caratheodory sense, that is $$ \forall S \subset \Omega \quad \mu(S) = \mu(S \cap E) + \mu(S \setminus E) \quad (*)$$ Then $E$ is also $\nu$-measurable in Caratheodory sense, that is $$ \forall S \subset \Omega \quad \nu(S) = \nu(S \cap E) + \nu(S \setminus E) \quad (\#)$$

My observation so far:

1. $\nu$ is an outer measure on $\Omega$.
2. if $\mu(E) = 0$, then $\nu(E) = 0$.
3. all open sets $\subset \Omega$ are $\nu$-measurable.
4. if $S$ in (#) is $\mu$-measurable, then (#) is true.

Edit: (extracted from the answer below): to extend (4) to all $S \subset \Omega$, use definition $(\$)$ to approximate $S$ with a $\mu$-measurable set $A_{\epsilon}$ with error $\epsilon$, then apply (4) to $A_{\epsilon}$ and let $\epsilon \to 0$: $$ \nu(S) - \epsilon \geq \nu(A_{\epsilon}) = \nu(A_{\epsilon} \cap E) + \nu(A_{\epsilon} \setminus E) \geq \nu(S \cap E) + \nu(S \setminus E) \quad \forall \epsilon > 0 $$

Answer 1

$\nu$ is a measure on the Lebesgue-measurable sets. This can be generalized to arbitrary measure spaces $(\Omega, \Sigma, \mu)$ and $\Sigma$-measurable non-negative functions $f$ on $\Omega$.

And if $\nu$ is a measure on $(\Omega, \Sigma)$, then $$\nu(E) := \inf \limits_{\Sigma \ni F \supseteq E} {\nu(F)}$$ defines an outer measure on $\Omega$. You have already shown this for your special case. What is left to show is that all elements of $\Sigma$ are $\nu$-measureable in the sense of Caratheodory: Let $S \subseteq \Omega$. If $\nu(S) =\infty$, there is nothing to prove. Otherwise $$\nu(S) \geq \nu(A_\varepsilon) - \varepsilon$$ for some $\Sigma$-measurable set $A_\varepsilon \supseteq S$. Now you can split $A_\varepsilon$ by $E$ and use the definition of $\nu$ again plus $\varepsilon \to 0$ to conclude.