Prove $\mu^*(A)=\nu(A)$ if there exists a cover $A\subset \cup_{n\geq1} B_n$ and $\mu^*(B_n)<\infty \;\forall n\geq 1$

Given $\mu : \mathscr{H} \to \mathbb{R}$ a pre-measure on $\mathscr{H}\subset X$ a semiring and $\mu^*$ the outer measure defined by $\mu$, $\mathscr{A}=\sigma(\mathscr{H})$ and $\nu$ a measure such that $\nu|_\mathscr{H}=\mu$, I want to show that
If for $A\in\mathscr{A}$ there exist $B_n\subset X$ such that $A\subset\cup_{n\geq 1}B_n$ and $\mu^*(B_n)<\infty$ for all $n\geq 1$, then $\mu^*(A)=\nu(A)$.

My attempt: I have already proven that $\mu^*(A)=\nu(A)$ for all $A\in\mathscr{A}$ that satisfy $\mu^*(A)<\infty$ but I am struggling to prove that this holds.

(Original title edited)

Lemma: If $B \subseteq X$ and $\mu^*(B) < \infty$, then there is $C \in \mathscr{A}=\sigma(\mathscr{H})$, such that $B \subseteq C$ and $\mu^*(C)=\mu^*(B)$.

Proof: Since $\mu^*(B) < \infty$ and $\mu^*(B) = \inf \left \{\sum_{n = 1}^\infty \mu(E_n): \forall n \geq 1, \ E_n \in \mathscr{H}, \ B \subseteq \bigcup_{n = 1}^\infty E_n \right\}$, we have, for each $k \in \Bbb N$,$k \geq 1$, there is a family of sets $\{E_{n,k}\}_{n \in \Bbb N}$ such that, for all $n \geq 1$, $E_{n,k} \in \mathscr{H}$, $B \subseteq \bigcup_{n = 1}^\infty E_{n,k}$ and

$$ \mu^*(B) \leq \sum_{n = 1}^\infty \mu(E_{n,k})\leq \mu^*(B) + \frac{1}{k} < \infty$$

In particular, for each $k \geq 1$, $\bigcup_{n=1}^\infty E_{n,k} \in\mathscr{A} $ and

$$ \mu^*(B) \leq \mu^*\left (\bigcup_{n=1}^\infty E_{n,k} \right) \leq\sum_{n = 1}^\infty \mu(E_{n,k})\leq \mu^*(B) + \frac{1}{k} < \infty$$

Now, for each $r \in \Bbb N$, $r\geq 1$, let $D_r= \bigcap_{k=1}^r\bigcup_{n=1}^\infty E_{n,k}$. It is immediate that, $D_r \in\mathscr{A} $, $B \subseteq D_r$ and

$$ \mu^*(B) \leq \mu^*(D_r) \leq \mu^*\left (\bigcup_{n=1}^\infty E_{n,r} \right) \leq\sum_{n = 1}^\infty \mu(E_{n,r})\leq \mu^*(B) + \frac{1}{r} < \infty$$

So, $\lim_{r \to \infty} D_r= \mu^*(B)$. On the other hand, $\{D_r\}_r$ is a decresing sequence of sets in $\mathscr{A}$ and $\mu^*$ restricted to $\mathscr{A}$ is a measure, so defining $C=\bigcap_{r=1}^\infty D_r $, we have that $C \in \mathscr{A}$, $B \subseteq C$ and $\mu^*(C) = \lim_{r \to \infty} D_r= \mu^*(B)$.

$\square$.

Suppose $A\in\mathscr{A}$ and there exist $B_n\subset X$ such that $A\subseteq\bigcup_{n\geq 1}B_n$ and $\mu^*(B_n)<\infty$ for all $n\geq 1$. By the lemma above, for each n\geq 1, there is $C_n \in \mathscr{A}$ such that $B_n \subseteq C_n$ and $\mu^*(C_n) = \mu^*(B_n) < \infty$. Let $F_1=C_1$ and for each $n \geq 2$, $F_n= C_n \setminus \bigcup_{i=1}^{n-1}C_i$. Then, for all $n\geq 1$, $F_n \in \mathscr{A}$, $\{F_n\}_{n\geq 1}$ is a family of disjoint sets, $\mu^*(F_n) \leq \mu^*(C_n) < \infty$, and $A\subseteq\bigcup_{n\geq 1}B_n \subseteq \bigcup_{n\geq 1}C_n =\bigcup_{n\geq 1}F_n $.

So, $A = \bigcup_{n\geq 1}A \cap F_n$, $\{A\cap F_n\}_{n\geq 1}$ is a family of disjoint sets in $\mathscr{A}$, and, for all $n\geq 1$, $\mu^*(A \cap F_n) < \infty$. Since, $\mu^*$ restricted to $\mathscr{A}$ is a measure,and you have already proved that $\mu^*(A)=\nu(A)$ for all $A\in\mathscr{A}$ that satisfy $\mu^*(A)<\infty$, we have $$ \mu^*(A) = \sum_{n=1}^\infty\mu^*(A \cap F_n) = \sum_{n=1}^\infty \nu(A \cap F_n)= \nu(A)$$