sum of a collection of truncated outer measures An outer measure on a set $X$ is a function $\mu:{\cal P}(X)\to[0,\infty]$ such that $\mu(\emptyset)=0$, $U\subseteq V$ implies $\mu(U)\le \mu(V)$, and $\mu(\bigcup_{n\in \Bbb N}U_n)\le\sum_{n\in\Bbb N}\mu(U_n)$ for any countable collection of sets $U_n\subseteq X$.
Given a $\sigma$-algebra $\Sigma$ on $X$ (a measurable space), in which the elements of $\Sigma$ are called measurable, the truncation of $\mu$ to $\Sigma$ is defined as $\mu_\Sigma(U)=\inf\{\mu(V)\mid V\supseteq U\wedge V\mbox{ is measurable}\}$. $\mu_\Sigma$ is an outer measure if $\mu$ is, and coincides with $\mu$ on $\Sigma$. (It is maximal among outer measures agreeing with $\mu$ on $\Sigma$.)
Given a collection of measures $\mu_i$ for $i\in I$, the sum outer measure is defined as $(\sum_i\mu_i)(U)=\sum_i\mu_i(U)$, where the infinite (possibly uncountable) sum over $I$ is the supremum of all finite partial sums. This is also an outer measure if the $\mu_i$ are.
Now, the question:

Let $(X,\Sigma)$ be a measurable space and let $(\mu_i)_{i\in I}$ be a collection of outer measures on $X$. Is $(\sum_i\mu_i)_\Sigma=\sum_i(\mu_i)_\Sigma$?

I have managed to prove this in the case when $I$ is finite, and one inequality is easy. If $V\supseteq U$ is measurable, then $\sum_i\mu_i(V)=\sum_i(\mu_i)_\Sigma(V)\ge \sum_i(\mu_i)_\Sigma(U)$ by monotonicity, so $\inf_V\sum_i\mu_i(V)=(\sum_i\mu_i)_\Sigma(U)\ge \sum_i(\mu_i)_\Sigma(U)$. Is the other direction true? Any references would be good as well - I haven't found this "truncation" definition in the literature (the name is my own) and I'd like to know what the usual terminology is here.
 A: Fix $U$, and let $V$ range over measurable supersets of $U$. The claim is that $$\inf_V\sum_i\mu_i(V)\le \sum_i\inf_V\mu_i(V).$$
Let $X=\omega_1\cup\{\infty\}$, and let a set be measurable if it is countable and does not contain $\infty$, or it is co-countable and contains $\infty$. Let $U=\{\infty\}$, and let $I=\omega_1$, where $\mu_i$ is the Dirac measure at $i\in\omega_1$ (i.e. $\mu_i(U)=1$ if $i\in U$, otherwise $0$).
If $V$ is a measurable superset of $U$, then it is co-countable, so $\sum_i\mu_i(V)=\sum_{i\in V}1=\infty$ (since $\omega_1$ is uncountable, so is $V$). Thus $\inf_V\sum_i\mu_i(V)=\infty$ as well.
But for any fixed $i\in\omega_1$, $\inf_V\mu_i(V)=0$ since $X-\{i\}$ is co-countable and omits $i$, so that $\mu_i(X-\{i\})=0$. Thus
$$\sum_i\inf_V\mu_i(V)=\sum_i0=0<\inf_V\sum_i\mu_i(V)=\inf_V|V|=\infty,$$
and the claimed inequality is false.

The claim is true when $I$ is countable. Given $\epsilon>0$, let $\epsilon_i>0$ be a sequence such that $\sum_i\epsilon_i=\epsilon$. Choose $V_i$ such that $\mu_i(V_i)\le \inf_V\mu_i(V)+\epsilon_i$. Then $\bigcap_{j\in I}V_j$ is also a measurable superset of $U$ since $I$ is countable, and:
\begin{align}
\inf_V\sum_i\mu_i(V)\le\sum_i\mu_i\big(\bigcap_jV_j\big)\le \sum_i\mu_i(V_i)&\le\sum_i\big[\inf_V\mu_i(V)+\epsilon_i\big]\\
&=\sum_i\inf_V\mu_i(V)+\epsilon.
\end{align}
Thus $\inf_V\sum_i\mu_i(V)=\sum_i\inf_V\mu_i(V)$.
