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Let $E$ be a measurable set, let $A$ be any set, then if $m^*$ is the outer measure function, $$m^*(A \cup E )=m^*(E)+m^*(A).$$

Is this true? By subadditivity, we have $m^*(A \cup E )\leq m^*(E)+m^*(A)$. I want to show that $m^*(A \cup E )\geq m^*(E)+m^*(A)$ so the equality holds. I thought of applying the definition of measurable set $E$ by $$m^*(A \cup E )\geq m^*((A\cup E)\cap E) + m^*((A\cup E)\cap E^c)$$ but I'm not getting the right result. Any hint?

Edit: if we add the hypothesis that $E$ and $A$ are disjoint, then by measurability of $E$, we can simplify the above inequality by: $$m^*((A\cup E)\cap E) + m^*((A\cup E)\cap E^c) = m^*(E) + m^*(A).$$ Since if $A \cap E =\emptyset $ then $A\subseteq E^c$. What about this?

Kurome
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  • What if $A=E $? – Eclipse Sun Apr 07 '16 at 15:59
  • Yep, seems like a counter example. This is false then? I might have misremembered/misread this question. – Kurome Apr 07 '16 at 16:03
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    This might be true if $A$ and $E$ are disjoint though. – Kurome Apr 07 '16 at 16:07
  • Sorry...misread the question. Is this what you're looking for? http://math.stackexchange.com/questions/469027/for-an-outer-measure-m-does-me-cup-ame-cap-a-mema-alw?rq=1 – Nicholas Stull Apr 07 '16 at 16:15
  • I don't think so. – Kurome Apr 07 '16 at 16:26
  • It is not true in general. Counter-example: as @EclipseSun pointed out take $A=E$. On the other, it is true if $A\cap E= \emptyset$ as the OP pointed out in the "edit" in the question. In fact, it is true if $m^(A\cap E)=0$, because then, since $E$ is measurable, $m^(A)=m^*(A\cap E^c)$. – Ramiro Apr 07 '16 at 20:15

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