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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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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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