# Show that for any $\alpha\in (0,\lambda(E))$, there exists a measurable set $F\subset E$ such that $\lambda(F)=\alpha$.

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?

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 $$|\phi (x)-\phi (y)| \leq |x-y|$$.
• 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! – user588903 Mar 6 at 0:06