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I have to prove that if $E$ is a set such that $\mu (E)=1$, then, there is a subset $E_t\subset E$ such that $\mu(E_t)=t$ for every $0\leq t\leq 1$

I think, that given a measurable set $E$ there is a Borel Set $F$ such that $\mu(F)=\mu(E)$ and then $F$ is union of disjoint that would be enough, but I am not sure about the last affirmation. Is it true that every Borel set a union of disjoint intervals?

(Here $\mu$ is the Lebesgue measure)

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Use continuity of the measure and consider $[-x,x]$ as $x$ grows. – Adam Hughes Jul 9 '14 at 7:13
This had been answered before, but I am unable to find a link. – copper.hat Jul 9 '14 at 7:19
In fact, if anyone's interested, for each $t$ such that $0 \leq t < 1,$ there is a perfect set $P_t$ such that $P_t \subseteq E$ and $\mu(P_t) = t.$ (I'm assuming we're working in the space ${\mathbb R}^{n}.)$ – Dave L. Renfro Jul 9 '14 at 14:22

Let $f(x) = \int_{(-\infty, x]} 1_E$. The function $f$ is continuous, $\lim_{x \to -\infty} = 0$ and $\lim_{x \to +\infty} = 1$. The intermediate value theorem does the rest.

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Thanks, now i see it, and with the help of this other question that i found… – Dimitri Jul 9 '14 at 7:35

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