Measure here is Jordan Measure.

$d(A,B)=\inf\{\vert x-y:x\in A,y\in B\}$

So I can show the first direction fairly easily because it doesn't use the condition.


Since $S_i$ are measurable there exists poly rectangles $P_i$ such that $\mu S_i\leq\vert P_i\vert\leq \mu (S_i)+\frac{\epsilon}{2}$. Since $S_i\subseteq P_i$, $S_1\cup S_2\subseteq P_1\cup P_2$

thus $\mu(S_1\cup S_2\leq \vert P_1\cup P_2\vert=\vert P_1\vert+\vert P_2\vert\leq \mu S_1+\mu S_2+\epsilon$

so $\mu (S_1\cup S_2)\leq \mu S_1 +\mu S_2$

But to show $\mu (S_1\cup S_2) \geq \mu S_1 + \mu S_2$ I'm not sure. Obviously since the distance between any 2 points between them is always positive they don't share any points. Which I believe should mean I can construct poly rectangles $P_i$ for each $S_i$ which have lengths less then $d(S_1,S_2)$ such that $P_1\cup P_2$ should still cover $S_1 \cup S_2$


For the other direction, let $\epsilon>0$ and choose $d(S_1,S_2)>\delta>0$. Now, take a finite covering of the union by intervals, i.e. $S:=S_1\cup S_1\subseteq \bigcup I_j$ such that $\sum |I_j|\le \mu^*(S)+\epsilon,$ and divide each $I_j$ into subintervals of length $<\delta.$

Each new interval can intersect at most one of the two sets $S_1, S_2$ so if we let the sets $J_1$ and $J_2$ contain the indices of the intervals in $I_j$ which intersect $S_1$ and $S_2,\ $ respectively, then

$$J_1\cap J_2=\emptyset,\ S_1 \subseteq \bigcup_{j∈J_1} I_j;\ S_2 \subseteq \bigcup_{j\in J_2} I_j $$

and therefore

$$\mu^*(S_1)+\mu^*(S_2)\le \sum_{j∈J_1} |I_j|+\sum_{j∈J_2} |I_j|=\sum |I_j|\le \mu^*(S)+\epsilon.$$


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