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I proved the following by contradiction, differently from how it is usually proven. I was told that the proof is missing cases but I don't understand why.

Let $\lambda$ denote Lebesgue measure. Let $E\subset\mathbb{R}$ be a measurable set with $0<\lambda(E)<\infty$. Show that for any $\alpha\in(0,\lambda(E))$, there exists a measurable set $F\subset E$ such that $\lambda(F)=\alpha$.

My proof: Assume instead that there exists some $\alpha\in(0,\lambda(E))$ such that for every $F\subset E$, we have $\lambda(F)\neq\alpha$. So we have two cases: Either $\alpha<\lambda(F)$ for every $F\subset E$, or $\alpha>\lambda(F)$ for every $F\subset E$.

If $\alpha<\lambda(F)$ for every $F\subset E$, then this must be true for $F=\emptyset$. Therefore, we get that $0<\alpha<\lambda(F)=\lambda(\emptyset)=0$, a contradiction.

Now, consider $\alpha>\lambda(F)$ for every $F\subset E$. We have a theorem that $\forall\epsilon>0$, there exists a closed set $F^*\subset E$ such that $\lambda(E\setminus F^*)<\epsilon$. From this, we get that $\lambda(E)-\epsilon<\lambda(F^*)$. Since $F^*\subset E$, then $\lambda(F^*)<\alpha$. So we get that $\lambda(E)-\epsilon<\lambda(F^*)<\alpha<\lambda(E)$. Since $\epsilon$ is arbitrary, then it must be true that $\lambda(F^*)=\alpha$, a contradiction.


Again, I was told that I'm missing two cases. But I don't understand why. Can someone explain?

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Your proof is wrong. You cannot say $\alpha <\lambda (F)$ for all $F \subset E$ or $\alpha >\lambda (F)$ for all $F \subset E$. Why cannot the first inequality hold for some $F$'s and the second for other $F$'s? For a corrcet proof define $\phi(x) =\lambda (E \cap (-\infty,x ))$ and use IVP. To show continuity of $\phi$ show that $|\phi (x)-\phi (y)| \leq |x-y|$.

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  • $\begingroup$ Ohh ok. I didn't consider that $\lambda(F)\neq\alpha$ could mean some $F$'s give $\lambda(F)<\alpha$ and others give the flipped inequality. Thank you! $\endgroup$ – user588903 Mar 6 at 0:06

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