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Is $\lambda ^{*} (A \setminus B) \leq \lambda^{*}(A)-\lambda^{*}(B)$ generally true? Where $\lambda^*$ is the Lebesgue outer measure, and I am assuming $B \subseteq A$.

It seems correct intuitively. Under which conditions does this happen? For all sets? Only Lebesgue measurable sets?

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    $\begingroup$ $\lambda^{*}(A \setminus B) \geq \lambda^{*}(A) - \lambda^{*}(B)$ is always true for any outer measure $\lambda^{*}$, but I'm not sure under what conditions the reverse inequality holds (except if the outer measure is actually a measure, in which case you have equality, at least if $B \subseteq A$). $\endgroup$ – layman Dec 11 '17 at 23:24
  • $\begingroup$ I did forget the $B \subseteq A$ part. $\endgroup$ – Bary12 Dec 11 '17 at 23:43
  • $\begingroup$ The problem is, in my opinion, phrased ambiguously when $\lambda^*(A)=\lambda^*(B)=\infty$. This might look unessential, but it is actually quite important. For instance, if I were to phrase it as "Which sets $A\supseteq B$ satisfy $\lambda^*(A)\ge \lambda^*(A\setminus B)+\lambda^*(B)$?", one could observe that this is satisfied as soon as $\lambda^*(A)=\infty$, reagardless of measurability of either (namely, a set $B$ may fail the measurability condition only when tested against a set of finite outer measure). $\endgroup$ – user228113 Dec 11 '17 at 23:53
  • $\begingroup$ I suggest you to read about non-measurable sets. $\endgroup$ – i707107 Dec 11 '17 at 23:55
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Consider the special case where $A$ is measurable. Then the right side of your inequality is $\lambda(A)-\lambda^*(B)$, which is the inner measure of $B$. So, unless $B$ is also measurable, your proposed inequality won't hold: Outer measure is $\leq$ inner measure only for measurable sets.

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