Let $\mu$ and $\lambda$ be two $\sigma$-finite premeasures on a ring $\mathcal{R}$. The induced outer measure on $\mu$ is
$$\mu^* :=\inf\left\{\sum_{n=1}^{\infty} \mu(A_n)\right\}$$
where $A_n \in \mathcal{R}$ and $A \subset \bigcup_{n=1}^{\infty} A_n$. The outer measure $\lambda^*$ induced by $\lambda$ is similarly defined. Further, let $\mathcal{M}$ and $\mathcal{L}$ be the set of $\mu^*$-measurable sets and $\lambda^*$-measurable sets, respectively, in the sense of the Caratheodory condition.
One can also show that $\mu + \lambda$ is a premeasure and let $\mathcal{N}$ be the set of all $(\mu + \lambda)^*$-measurable sets.
How to show that
$$\mathcal{M} \bigcap \mathcal{L} = \mathcal{N}?$$
For the $\subset$ direction, I just used the Caratheodory condition, so I think for the $\supset$ direction, I should use the $\sigma$-finiteness of $\mu$ and $\lambda$; that is, there exists $\{A_n\}_{n=1}^{\infty} \subset \mathcal{R}$ such that $\mu(A_n) \le \infty$ for every $n \in \mathbb{N}$ and $X = \bigcup_{n=1}^{\infty} A_n$, and similarly for $\lambda$.
I am thinking of proving that $\mu^*$ and $\lambda^*$ are also $\sigma$-finite on $\mathcal{M}$ and $\mathcal{L}$, respectively. But I'm not sure how could this be helpful. Do you have an idea on how I could continue?