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I need to show that for every $\epsilon>0$ there exists a dense open subset $V$ of $\mathbb{R}$ such $m(V)=\epsilon$. Here, $m$ is the Lebesgue measure.

Prior to this, I showed that for every $\epsilon>0$ there exists a dense open subset $U$ of $\mathbb{R}$ such $m(U) \le \epsilon$.

I showed the latter statement by taking $U$ to be the union of the open intervals $I_{n}=(a_{n}-\frac{\epsilon}{2^{n+2}}, a_{n}+\frac{\epsilon}{2^{n+2}})$.

Then $m(U)=m(\cup I_{n}) \le \sum m(O_{n})=\frac{\epsilon}{2}<\epsilon$.

However, I don't know how to prove the first statement. Any hints?

KCd
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Live Free or π Hard
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  • What is the definition of "a dense open subset of $\mathbb{R}$"? – Evan Aad Oct 04 '16 at 17:57
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    My background isn't analysis, but from what I understand, the definition of a dense open subset of $\mathbb{R}$ is that it's closure is all of $\mathbb{R}$. I think that if I could shrink my original set of intervals, so that they are disjoint, I could have equality, but I'm not sure how. – Live Free or π Hard Oct 04 '16 at 18:11
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    Use your construction: for each $\epsilon$, $\epsilon/2<m(U_\epsilon)<\epsilon$, and $m(U_\epsilon)<m(U_\eta)<m(U_\epsilon)+\eta-\epsilon$ for each $\epsilon<\eta$, hence the function $\epsilon\mapsto m(U_\epsilon)$ is continuous and, for every positive $\epsilon$, there exists $\eta$ in $(\epsilon,2\epsilon)$ such that $m(U_\eta)=\epsilon$. – Did Oct 04 '16 at 18:45

3 Answers3

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Let $G$ be a dense, open subset of $\mathbb{R}$, whose Lebesgue measure is positive and finite. Define a function $f:\mathbb{R}\rightarrow\mathbb{R}$ as follows. For every $x\in\mathbb{R}$, $$ f(x) := \frac{\varepsilon}{m(G)}x. $$ Since $f$ is linear, $m\left(f(G)\right) = \frac{\varepsilon}{m(G)} m(G) = \varepsilon$ (see here). Since $f$ is continuous, $f(G)$ is dense in $\mathbb{R}$ (see here). Since $f$ is invertible, and its inverse is again continuous, $f(G)$ is open.

Community
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Evan Aad
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Fix an enumeration $\{q_n\}_{n=1}^{\infty}$ of the rationals, and for each real number $t\geq 0$, define $$ U_t=\bigcup_{n=1}^{\infty}\Big(q_n-\frac{t}{2^n},q_n+\frac{t}{2^n}\Big) $$ Then $U_0=\emptyset$, and for $t>0$ $U_t$ is a dense open subset of $\mathbb{R}$. Note also that $$ t\leq m(U_t)\leq t\sum_{n=1}^{\infty}2^{1-n}=2t $$ where $m$ is Lebesgue measure.

Define a function $f(t)=m(U_t)$ for $t\geq 0$. Then $f(0)=0$ and $f(t)\to\infty$ as $t\to\infty$. One can show that $f(t)$ is continuous, hence for all $a>0$ there exists $t>0$ such that $m(U_t)=a$ by the intermediate value theorem.

carmichael561
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Once you have open dense sets of arbitrarily small measures, the remaining solution is quite straightforward.

Let $U$ be an open dense set of measure $\varepsilon'\leq\varepsilon$.

The function $f\colon {\bf R}_{\geq 0}\to {\bf R}_{\geq 0}$ defined by $f(t)=m(U\cup (0,t))$ is obviously Lipschitz, and hence continuous, as well as unbounded, whence we easily get that it is onto $[\varepsilon',\infty)$. Then $U'=U\cup (0,t_0)$, where $t_0\in f^{-1}[\varepsilon]$, has the desired property by definition of $f$.

tomasz
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