Given a set $X$, a collection of subsets $K\subseteq \wp(X)$ such that $\varnothing\in K$ and a function $m:K\to[0,\infty]$ such that $m(\varnothing)=0$, we define the function $\lambda:\wp(X)\to [0,\infty]$ by $$\lambda(A)=\inf\left\{\sum_{n\in\mathbb{N}}m(B_n): A\subseteq \bigcup_{n\in\mathbb{N}}B_n\text{ and }\{B_n\}_{n\in\mathbb{N}}\subseteq K \right\}.$$
Then, it is not difficult to see that this is in fact an outer measure on $X$.
It follows from the definition, that given any $A\in K$, $\lambda(A)\leq m(A)$. What I'm trying to figure out is whether or not $\lambda|_{K}=m$.
If $m$ satisfies that for any $A\in K$ and any $\{B_n\}_{n\in\mathbb{N}}\subseteq K$ such that $A\subseteq \bigcup_{n\in\mathbb{N}}B_n$, $m(A)\leq \sum_{n\in\mathbb{N}}m(B_n)$, we would have the result, since in this case $m(A)$ would be a lower bound for the set which $\lambda(A)$ is an infimum of.
Any idea of how to prove that in fact $\lambda|_{K}=m$ or of how to find a counterexample is greatly appreciated.
Thanks in advance.
It is not true in general. Let $X$ be a finite set. Let $K$ be the powerset of $X$. Set $m(X) = 1$, and $m(E) = 0$ for all other sets $E \in K$. Then $\lambda(X) = 0$ while $m(X) = 1$.
If you assume that $K$ is an algebra of subsets of $X$ and that $m$ is a premeasure on $K$, then we do have $\lambda = m$ on $K$. This is part of Caratheodory's theorem.