Infimum measure of unions Let $X$ be a metric space with $f:\mathcal{P}(X)\rightarrow[0,\infty]$, such that $f(\emptyset)=0.$ And let $\hat{E}_{a}=\{E\in\mathcal{P}(X):\text{diam}(E)\leq a\}$.
Now we define
$$m_a(Y)=\inf\{\sum_{n=1}^\infty f(E_n):E_n\in \hat{E}_a, Y\subset\cup_{n=1}^\infty E_n\}$$
and
$$m(Y)=\lim_{a\rightarrow 0}m_a(Y).$$

How do I prove that $m(\cup_{n=1}^\infty U_n)\leq\sum_{n=1}^\infty m(U_n)?$

What I thought of:
First let $U=\cup_{n=1}^\infty U_n$ and 
$$m_a(U_n)=\inf\{\sum_{k=1}^\infty f(E_k^n):E_k^n\in \hat{E}_a, U_n\subset\cup_{k=1}^\infty E_k^n\}.$$
Then
$$m_a(U)=\inf\{\sum_{i=1}^\infty f(A^i):A^i\in \hat{E}_a, U=\cup_{n=1}^\infty U_n\subset\cup_{i=1}^\infty A^i\},$$
since $U_n\subset\cup_{k=1}^\infty E_k^n$, we can take as $A_i=\cup_{k=1}^\infty E_k^i$.
However does it still hold that $\cup_{k=1}^\infty E_k^n\in \hat{E}_a$? And how would I continue?
 A: Notice first that each $m_a$ is an outer measure, hence satisfies
$$ m_a\left(\bigcup_{n=1}^{\infty} U_n \right) \leq \sum_{n=1}^{\infty} m_a (U_n). \tag{*}$$
Moreover, for each fixed $U$ the function $a \mapsto m_a(U)$ is non-increasing. Consequently we always have $m(U) \geq m_a(U)$ for all $a > 0$. So
$$ m_a\left(\bigcup_{n=1}^{\infty} U_n \right) \leq \sum_{n=1}^{\infty} m (U_n) $$
and taking $a \to 0^+$ gives the desired claim.

If you need to prove $\text{(*)}$ (and I suspect so, based on your thought), you may rely on the following very standard result.

Lemma. Let $X$ be a non-empty set and assume that
  
  
*
  
*$\mathcal{S}$ is a cover of $X$ containing $\varnothing$, and
  
*$f : \mathcal{S} \to [0, \infty]$ is a function satisfying $f(\varnothing) = 0$.
  
  
  Then the set function $\mu^{\star} : \mathcal{P}(X) \to [0, \infty]$ defined by
  $$ \mu^{\star}(E) = \inf\bigg\{ \sum_{n=1}^{\infty} f(E_n) : \{E_n\} \subseteq \mathcal{S} \text{ and } E \subseteq \bigcup_{n=1}^{\infty} E_n \bigg\} $$
  is an outer measure on $X$.

Proof. $\mu^{\star}(\varnothing) = 0$ and the monotonicity is trivial from the definition of $\mu^{\star}$. So it suffices to check the countable subadditivity.
To this end, let $E = \bigcup_{i=1}^{\infty} E_i$ and for each $\epsilon > 0$ and $i$, pick $\{E_{i,n}\}_{i=1}^{\infty}$ such that $E_i \subseteq \bigcup_{n=1}^{\infty} E_{i,n}$ and $\sum_{n=1}^{\infty} f(E_{i,n}) \leq \mu^{\star}(E_i) + \epsilon 2^{-i}$. Then $\{ E_{i,n} \}_{i,n=1}^{\infty} \subseteq \mathcal{S}$ satisfies $E \subseteq \bigcup_{i=1}^{\infty}\bigcup_{n=1}^{\infty} E_{i,n}$ and
$$ \mu^{\star}(E)
\leq \sum_{i=1}^{\infty}\sum_{n=1}^{\infty} f(E_{i,n})
\leq \sum_{i=1}^{\infty}\mu^{\star}(E_i) + \epsilon. $$
Letting $\epsilon \to 0^+$ proves the desired result. ////
