real analysis - Finite positive measure - integrals over subsets

Let $\mu$ be a real measure on a space $(S,\Sigma)$. Define $\nu: \Sigma \to [0, \infty)$ by
\begin{align}

\nu(E) = \sup\{\mu(F): F \in \Sigma, F \subseteq E, \mu(F)\geq 0 \}.
\end{align}
I want to show that $\nu$ is a finite positive measure.

Therefore, $\nu(\emptyset)= \sup\{0\}=0$. And I want to show that $\nu(\sqcup_{n \in \mathbb{N}} E_n) = \sum_{n=1}^\infty \nu(E_n) $ using the monotone convergence theorem.

I can propose that the disjoint union of $E_n$'s form $E$, i.e. $\sqcup_{n \in \mathbb{N}} E_n = E$. So,
\begin{align}
\nu(\sqcup_{n \in \mathbb{N}} E_n) = \nu(E).
\end{align}

Now, I want to write $E$ as a sum of events which are a monotone increasing sequence such that the monotone converging theorem can be applied. However, I can't come up with such a sequence.

Answer

The idea is the same as the proof for positive measures. Note that the restriction $\mu(F)\ge0$ is redundant in the definition of $\nu$. There always exists $F\subseteq E$ such that $\mu(F)\ge 0$, e.g., $F=\emptyset$, so subsets with negative measure will be "automatically" ignored when taking supremum.

We first prove the sub-additivity of $\nu$. Suppose $A\cap B=\emptyset$. For any $F\subseteq A\cup B$, we have $\mu(F)=\mu(F\cap A)+\mu(F\cap B)\le\nu(A)+\nu(B)$. The sub-additivity follows.

Then we prove $\sum_{n=1}^\infty\nu(E_n)\le\nu(\sqcup_{n=1}^\infty E_n)$. By the definition of supremum, for each $E_n$ there is some subset $F_n\subseteq E_n$ such that $\nu(E_n)-\mu(F_n)<\epsilon\cdot 2^{-n}$. Hence
$$\sum_{n=1}^\infty\nu(E_n)<\epsilon+\sum_{n=1}^\infty\mu(F_n)=\epsilon+\mu(\bigsqcup_{n=1}^\infty F_n)$$
Since $\sqcup F_n\subseteq\sqcup E_n$, it's immediate that $\mu(\sqcup F_n)\le\nu(\sqcup E_n)$. Now we're done.