Prove directly from the definition of Outer Measure $m^{*}$ that if $m^{*}(A) = 0$, then $m^{*}(A \cup B)=m^{*}(B)$ (this is actually an exercise from Royden, but our professor wants us to use the definition of Outer Measure (a function $m^{*}: \{\text{subsets of}\,\mathbb{R}\} \to \mathbb{R}_{\geq 0} \cup \{\infty \}$ defined as:
$m^{*}(A) = \inf \left \{\displaystyle \sum_{k=1}^{\infty} l(I_{k})\, \vert I_{1},I_{2},\cdots \text{open bounded intervals such that}\, A \subseteq \cup_{k=1}^{\infty} I_{k} \right\}$ (where $l(I_{k})$ is the length of the $k$th interval)).
Now, let $m^{*}(B) = \inf \left \{\displaystyle \sum_{k=1}^{\infty} l(J_{k})\, \vert J_{1},J_{2},\cdots \text{open bounded intervals such that}\, B \subseteq \cup_{k=1}^{\infty} J_{k} \right\}$ and $m^{*}(A \cup B) = \inf \left \{\displaystyle \sum_{k=1}^{\infty} l(M_{k})\, \vert M_{1},M_{2},\cdots \text{open bounded intervals such that}\, A \cup B \subseteq \cup_{k=1}^{\infty} M_{k} \right\}$
The way I think the proof should go is the following:
$m^{*}(A \cup B) = \inf \sum_{k=1}^{\infty} l(M_{k}) = \inf \sum_{k=1}^{\infty}l(I_{k}) + \inf \sum_{k=1}^{\infty}l(J_{k}) = 0 + m^{*}(B) = m^{*}(B)$.
However, in order to do this, I would (1) need to show that $\cup_{k=1}^{\infty} M_{k} = (\cup_{k=1}^{\infty}I_{k})\cup (\cup_{k=1}^{\infty}J_{k})$ and (2) that this implies that $\inf \sum_{k=1}^{\infty} l(M_{k}) = \inf\sum_{k=1}^{\infty}l(I_{k}) + \inf\sum_{k=1}^{\infty}l(J_{k})$.
It's frustrating, too, because this is supposed to be an easy exercise, and intuitively, it makes sense that the absolute minimum needed to cover $A \cup B$ should also be the absolute minimum needed to cover $A$ plus the absolute minimum needed to cover $B$, but I need to show this in a mathematically rigorous way, and can't just assert things without being able to mathematically explain them.
If someone could please help me fill in the justification gaps in this proof and/or fix it if it's incorrect, it would be very much appreciated, and would definitely help me in my attempts to tackle the harder problems.