I have a problem:
Suppose $m$ is a measure on algebra $\mathcal{C}$. Then
$$\mu^*(A)= \inf\left\{\sum_{i=1}^\infty m(P_i)\mid A\subset\bigcup_{i=1}^{\infty}P_i, P_i\in\mathcal{C}\right\}$$ is outer measure and $\mu^*(A)= m(A)$ if $A\in \mathcal{C}$.
We use Carathedory's Theorem to create a measure $\mu$ on $\sigma$- algebra $\mathcal{L}\supset\mathcal{F}(\mathcal{C})\supset\mathcal{C}.$ Assume $A \in\mathcal{C}$ and use the definition, we find for any $k=1,2,...,\mathcal{P}_{ik}\in\mathcal{C}$ such that $A\subset\bigcup_{i=1}^{\infty}P_{ik}$ and $ \sum_{i=1}^{\infty}m(P_{ik})\le \mu^*(A)+1/k=\mu(A)+1/k.$
Let $B= \bigcap_{k=1}^{\infty}\bigcup_{i=1}^{\infty}P_{ik}$, we have $A\subset{B}, B\in\mathcal{F}(\mathcal{C})$ and $\mu(B)\le\mu(A)$.
Since $A\subset{B}$, thus $\mu(B)=\mu(A)$.
I want to show that $\mu(B\setminus A)=0$.
Your deduction is invalid when $\mu(A)=+\infty$ and when $\mu(A)<+\infty$ we have $$\mu(B)=\mu(A)+\mu(B\setminus A)$$ to obtain $$\mu(B\setminus A)=0.$$