# Suppose $A \subset [0,1]$, $B \subset [1,2]$. We do not assume they are Lebesgue measurable?

Suppose $A \subset [0,1]$, $B \subset [1,2]$ (we do not assume they are Lebesgue measurable). Show that $m^{*}(A \cup B) = m^{*}(A) + m^{*}(B)$.

Just learned this. Not sure about it. I used $m^{*}(A \cup B)-m^{*}(B) \le m^{*}(A)$. Please help with proof.

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This problem is actually very easy if you understand the definition of $m^{*}$ (I think). Do you? If not, what don't you understand? – Quinn Culver Mar 18 at 4:54
I agree with user67236 who says that we should help user67236. – Quinn Culver Mar 18 at 5:13
Here is what I know so far because we just learned some of it not much. If A is measurable and B is any set with A intersect B  = empty set. Letting A intersect B = empty set (A intersect B)intersect A is equal to A+(A union B)-A = B. That is the only thing I learned so far in this topic. – 9959 Mar 18 at 5:38
You said $A$ and $B$ are not assumed measurable yet used some fact that begins "If $A$ is measurable". This seems like a problem, doesn't it? And I'm sure you've learned more in this topic; e.g. the definition of $m^*$. Do you have the definition of $m^{*}$? – Quinn Culver Mar 18 at 5:57
That is the problem. It was given as trick question that was not learned yet. That is why I used that statement. Thanks for seeing that. I think if they are non measurable,we could have A intersect B equal the empty set. m*(A union B) does not = m*(A)+m*(B). – 9959 Mar 18 at 6:04
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## 2 Answers

I assume $m^*$ denotes the outer Lebesgue Measure in $\mathbb{R}$. In this case, we already know from the definition that $$m^*(A \cup B) \leq m^*(A) + m^*(B).$$ Hence it suffices to show the inequality "$\geq$". By definition $$m^*(A) = \inf\{ \sum_{k=1}^\infty l(I_k): I_k \text{ open Intervalls and } A\subset\bigcup_{k=1}^\infty I_k\},$$ where for an interval $I = (a,b)$ we denote by $l(I) := b-a$ the length of the interval $I$. Now let $I_k$ be arbitrary open Intervalls such that $A\cup B\subset\bigcup_{k=1}^\infty I_k$. Without loss of generality $\sum_{k=1}^\infty l(I_k) < \infty$. Then we have $A\subset\bigcup_{k=1}^\infty (I_k\cap[0,1])$ and $B\subset\bigcup_{k=1}^\infty (I_k\cap[1,2])$ respectively. Furthermore $$\sum_{k=1}^\infty l(I_k) \geq \sum_{k=1}^\infty l(I_k \cap [0,1]) + \sum_{k=1}^\infty l(I_k \cap [1,2]).$$ It follows from here that $$\begin{align} m^*(A\cup B) &= \inf\{ \sum_{k=1}^\infty l(I_k): I_k \text{ open Intervalls and } A \cup B \subset\bigcup_{k=1}^\infty I_k\} \\ &\geq \inf\{ \sum_{k=1}^\infty l(I_k \cap [0,1]): I_k \text{ open Intervalls and } A\cup B\subset\bigcup_{k=1}^\infty I_k\} \\ &\quad+ \inf\{ \sum_{k=1}^\infty l(I_k \cap [1,2]): I_k \text{ open Intervalls and } A\cup B\subset\bigcup_{k=1}^\infty I_k\}\\ &\geq m^*(A) + m^*(B) \end{align}$$

Please note, that the proof heavily relies on the fact, that $I_k \cap [0,1]$ is again an interval. The separation of $A$ and $B$ into disjoint intervals is the crucial assumption. Thus, a necessary condition for the outer Lebesgue measure to have strict subadditivity for two disjoint sets $A$ and $B$ is, that for any two intervals $I$ and $J$ the intersection $I \cap J$ is not a null set (in this case empty set or a point) whenever $A\subset I$ and $B\subset J$.

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The set $E=[0,1]$ is Lebesuge measurable. According to the definition of measurability, $$m^*(A \cup B)=m^*((A \cup B) \cap E)+m^*((A \cup B) \cap E^c)$$

Simplifying a little:

$(A \cup B) \cap E=(A \cap E) \cup (B \cap E)=A \cup(B \cap E)$

$B \subset [1,2]$ implies $B \cap E$ has at most one point, and therefore $m^*((A \cup B) \cap E)=m^*(A)$

Similarly, you can show that $m^*((A \cup B) \cap E^c)=m^*(B)$ which proves what's required.

I've used this known result: If $F$ is any set with $m^*(F)=0$, then $m^*(E \cup F)=m^*(E)$ for any $E \subseteq \mathbb R$

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