Show that subspace measure is a measure Let $(X, \Sigma, \mu)$ be a measure space and $D \subseteq X$.

Theorem:
Let
  $$ \mu_D(S) = \inf\{\mu(U)\mid U \in \Sigma, S \subseteq U\}: \Sigma_D \to [0,\infty]$$
  where $\Sigma_D = \{U \cap D\mid U \in \Sigma\}$
  Then $(D, \Sigma_D, \mu_D)$ is a measure space.

I am able to show that $\Sigma_D$ is a $\sigma$-algebra on $D$, but I am stuck while showing that $\mu_D$ is a measure. I tried the following:
Let $S, T\in \Sigma_D$ be disjoint sets. Then somehow I need to show that
$$\inf\{\mu(U)\mid U \in \Sigma, S \cup T \subseteq U\}
= \inf\{\mu(U) + \mu(V)\mid
     U, V \in \Sigma \land U, V \text{ are disjoint} \land
     S \subseteq U, T \subseteq V\}$$
However, that introduces a problem: Disjoint sets have disjoint supersets. I know that this is not always true, so what is incorrect here?
 A: Let $\{ A_{i}\}_{i=0}^{\infty} \subseteq \Sigma_{D}$ be pairwise disjoint. We need to show that 
$$\mu_{D}\left(\bigcup_{i=0}^{\infty} A_{i}\right) = \sum_{i=0}^{\infty} \mu_{D}(A_{i}).$$
Showing that $\mu_{D}(\cup A_{i}) \leq \sum \mu_{D}(A_{i})$ is fairly trivial: similar to any subadditivity proof, and using the subadditivity of $\mu$. Thus it remains to show that $\mu_{D}(\cup A_{i}) \geq \sum \mu_{D}(A_{i})$.
Well, let $\epsilon > 0$. Then there exists some $B \in \Sigma$ such that $B \supseteq \cup_{i=0}^{\infty} A_{i}$ and $\mu(B) \leq \mu_{D}(\cup_{i=0}^{\infty}A_{i}) + \epsilon$. By definition of $\Sigma_{D}$, for each $i \geq 0$, there exists some $B_{i} \in \Sigma$ such that $B_{i} \cap D = A_{i}$. Since the $A_{i}$'s are pairwise disjoint, $\{B_{i} \cap D\}_{i=0}^{\infty}$ is a pairwise disjoint sequence of sets. Define $$B_{i}^{*} := B \cap \left(B_{i} - \bigcup_{j\neq i}B_{j}\right), \ i \in \mathbb{N}.$$
Then $B_{i}^{*} \supseteq A_{i}$ and $\{ B_{i}^{*}\}_{i=0}^{\infty} \subseteq \Sigma$ is pairwise disjoint. Therefore,
$$\sum_{i=0}^{\infty} \mu_{D}(A_{i}) \leq \sum_{i=0}^{\infty} \mu(B_{i}^{*}) = \mu\left(\cup_{i=0}^{\infty} B_{i}^{*}\right) \leq \mu(B) \leq \mu_{D}\left(\cup_{i=0}^{\infty} A_{i}\right) + \epsilon.$$
Since $\epsilon > 0$ was arbitrary, $\sum_{i=0}^{\infty}\mu_{D}(A_{i}) \leq \mu_{D}\left(\cup_{i=0}^{\infty}A_{i}\right)$.
