Show that for every $\epsilon>0$ there exists a dense open subset $V$ of $\mathbb{R}$ such $m(V) = \epsilon$. 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?
 A: 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.
A: 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.
A: 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 $f$.
