# Let $S_1,S_2$ be measurable with $d(S_1,S_2)>0$. Show that $\mu(S_1\cup S_2)=\mu S_1+\mu S_2$

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.

$$\leq$$

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$$
$$\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.$$