Why my logic is flawed in this reasoning:-
If A is measurable we have Sup $\lambda (F)$ = inf $\lambda (G)$ for all such G and F, therefore we can always find such F and G which are $\epsilon$ / 2 close to $\lambda (A)$.
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What sort of set is $F$ supposed to be? My guess is compact, but that's just a guess until you say so. Why do you say there is a flaw in your reasoning? Gven $\epsilon>0$ what you say is exactly right. Now, in the title you say "there exist $F$ and $G$ such that for every $\epsilon>0$...", which is simply wrong; what's true is that "for every $\epsilon>0$ there exist $F$ and $G$ such that...". I have no idea whether that answers your question, because it's not very clear what you're asking. – David C. Ullrich May 08 '16 at 16:42
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Oops. Exactly right for sets of finite measure, that is. – David C. Ullrich May 08 '16 at 16:45
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The way you stated the problem, $F$ and $G$ are not allowed to depend on $\epsilon$; in fact, $\forall ε>0, λ(G ∖ F) < ε$ implies $\lambda(G ∖ F) = 0$. That can't be true: for example, if $A = \{ 0 \}$ any open set containing $A$ has to have strictly positive measure.
Daniel Schepler
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