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Please excuse the question, I didn't know how to put it more succinctly. Feel free to edit

Say I have a set $X$ and $\mathcal{A}\subseteq \mathcal{P}(X)$ and $\emptyset, X \in \mathcal{A}$. Let $\rho:\mathcal{A} \rightarrow [0,\infty]$ be a mapping with $\rho(\emptyset)=0$. Then this induces an outer measure ( this is given ):

$$ \mu^*:=\inf\{\sum\limits_{k\in \mathbb{N}}\rho(A_k)|(A_k)_{k \in \mathbb{N} }\subseteq \mathcal{A} \textit{ and } A \subseteq \cup_{k \in \mathbb{N}}A_k\} $$

Let's call the above set of the sums $M_A$. I'd like to show that $\rho \geq \mu^*$ ( not sure if this is true ).

Is it as simple as saying for any $A \in \mathcal{A}$,

$$ \mu^*(A)= \inf M_A $$ and because $A \subseteq A$, we have $A \in M_A$ and therefore

$$ \mu^*(A)= \inf M_A \leq \rho(A) $$ ?

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Yes, that's correct, up to some minor imprecisions in your wording. Note that you've defined $M_A$ as a set of numbers, not a set of sets, so it's not literally true that $A\in M_A$. Instead, you can take $A_0=A$ and $A_k=\emptyset$ for $k>0$ to get a sequence $(A_k)$ as in the definition of $\mu^*(A)$, so that $\sum_k \rho(A_k)=\rho(A)\in M_A$ so $\inf M_A\leq \rho(A)$.

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