I am trying to solve the exercise below. I have managed to prove that $\mu$ is an outer measure following the definition of outer measures ($\mu(\emptyset)=0$, monotonicity, $\sigma$-subadditivity). However, I am struggling to find $\mathcal{M}^*$, the collection of all $\mu$-measurable sets. It is clear that $\emptyset, X \in \mathcal{M}^*$ but I can't find more sets $A$, which satisfy $ \forall E \subseteq X \; \; \mu(E) = \mu(E \cap A) +\mu(E \cap A^c)$. I also managed to prove that if an outer measure is finitely additive, it is a measure in general, but I can't find the sufficient and necessary conditions to prove that $\mu$ is a measure.
Exercise:
If $X \ne \emptyset$ and $\mu$ is a set function defined on $\mathcal{P}(X)$ as \begin{equation} \mu(X) =
\begin{cases}
\text{Card}(E), & \text{Card}(E) < \aleph_0\\
\infty, & \text{Card(E)} \ge \aleph _0
\end{cases} \end{equation}
a) Prove that $\mu$ is an outer measure and find $\mathcal{M}^*$.
b) Prove that if an outer measure is finitely additive, it is also a measure. Find the necessary and sufficient conditions under which the outer measure $\mu$ is also a measure.
All subsets $A\subseteq X$ are $\mu$-measurable. If $E\subseteq X$ then $E=(E\cap A)\sqcup(E\setminus A)$. The cardinality of a disjoint union is always the sum of cardinalities, whether they're finite or not $$ \operatorname{Card}(E)=\operatorname{Card}(E\cap A)+\operatorname{Card}(E\setminus A)\tag{1} $$
If $E$ is finite, then all numbers in $(1)$ are finite and you get the conclusion $\mu(E)=\mu(E\cap A)+\mu(E\setminus A)$.
If $E$ is infinite then $\mu(E)=+\infty$. Since an infinite set cannot be a union of two disjoint finite sets, either $E\cap A$ or $E\setminus A$ is an infinite set. Therefore either $\mu(E\cap A)=+\infty$ or $\mu(E\setminus A)=+\infty$ (or both). Either way $\mu(E)=\mu(E\cap A)+\mu(E\setminus A)$.