For a measure zero set $A$, the union $A\cup B$ has zero measure if and only if $B$ does Definition: A set $A$ has measure $0$ iff $\forall \epsilon > 0, \exists$ system of intervals $(I_\tau): A \subseteq \cup_\tau (I_\tau), 0 \leq \sum_\tau (\operatorname{length}(I_\tau)) < \epsilon$.
Using this definition, prove that the following is true: Let $A$ be a set with measure $0$. Then the set $B$ has measure $0$ iff the set $A \cup B$ has measure $0$.

My thoughts: It seems obvious and trivial, which is actually the difficulty of it for me. The backward direction is easy, I think. If $A \cup B$ is contained in a system of intervals with total length less than $\epsilon$ (for some given $\epsilon$), then clearly $B$ is too (since $B \subseteq A \cup B$). But this is true of $\forall \epsilon > 0$, so we are done. In the forward direction, I would attempt the following: $\forall \eta > 0$, $A$ and $B$ each can be covered by systems of intervals of total length less than $\eta$, by assumption. If $A$ and $B$ are utterly/maximally disjoint, then $A \cup B$ can be covered by the union of their respective interval systems, which may also be disjoint (or not). If these systems are disjoint, then their union has total length less than $2 \eta$ (how do I prove it?). So, if we want to have the system of intervals covering $A \cup B$ (which clearly can be the union of the systems of intervals covering $A$ and $B$ separately and respectively) have a total length less than some given $\epsilon > 0$, we demand that $\eta < \frac{\epsilon}{2}$. But this works $\forall \epsilon > 0$ because $0 < \eta < \frac{\epsilon}{2}$ was arbitrary too (because each of $A$ and $B$ may be covered by satisfactory systems of intervals for any such $\eta$, by the definition of "measure 0" and by assumption). Done?

Yes, this proof works. The notes I would have towards it are:
It should be fairly obvious that, the total length in the union of two systems of intervals is exactly the sum of the total length in each interval - after all, those are both arising from absolutely convergent sums, so summing the results is the same as summing both sequences together.
Moreover, you don't need to fuss about whether a system of intervals is disjoint or not; we're only concerned about the sum of the lengths of the intervals, so we're not actually speaking about what set the intervals cover, but directly about the set of intervals. So, for instance, we'd probably say that the system $\{(0,1),(0,2)\}$ has length $1+2=3$ despite the fact that the union of those intervals is $(0,2)$, with length only $2$. Since the measure is defined as the infimum of the length of all systems of intervals covering the set, we would never actually worry about clearly suboptimal cases like this.
