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Any set $E$ of outer measure zero ($m^{*}(E) = 0$) is measurable.

My solution

Since $A\cap E\subseteq E$ and the outer measure is monotonic, we have that $0\leq m^{*}(A\cap E) \leq m^{*}(E) = 0$.

On the other hand, we have that $A = (A\cap E)\cup(A\cap E^{c})$

Consequently, it results that \begin{align*} m^{*}(A) \leq m^{*}(A\cap E) + m^{*}(A\cap E^{c}) = m^{*}(A\cap E^{c}) \end{align*}

Finally, since $A\cap E^{c}\subseteq A$, we have that $m^{*}(A\cap E^{c})\leq m^{*}(A)$, and the proposed claim holds: \begin{align*} m^{*}(A) = m^{*}(A\cap E^{c}) = 0 + m^{*}(A\cap E^{c}) = m^{*}(A\cap E) + m^{*}(A\cap E^{c}) \end{align*}

Is there another way to approach it or is it standard?

user0102
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1 Answers1

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This all looks fine to me so I would think it would be more reader taste. You could briefly justify why $m^*(A \cap E)=0$, i.e. $0 \leq m^*(A \cap E) \leq m^*(E)=0$. But some readers would find this wildly unnecessary. Or for instance, you could get rid of the middle equation and simply include your comment on $m^*(A \cap E^c) \leq m^*(A)$ and then $$ 0 \leq m^*(A \cap E^c) \leq m^*(A)= m^*(A \cap E) + m^*(A \cap E^c)= m^*(A \cap E^c) $$ It's now more a matter of the exposition and level of detail. For instance, here you can see a proof that is essentially just one line.

mathematics2x2life
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