The following is an exercise from Rudin's Real and Complex Analysis (where $m$ denotes Lebesgue measure):
Exercise 2.7: If $0<\varepsilon<1$, construct an open set $E\subseteq[0,1]$ which is dense in $[0,1]$, such that $m(E)=\varepsilon$.
Here is my solution to this exercise; a previous exercise was to show that the Cantor middle-thirds set had Lebesgue measure zero, and my solution is a variant of how I solved that problem.
Proof: Let $A_0=[0,1]$, and for each $A_n$, inductively define $A_{n+1}$ by removing the central interval of length $\varepsilon^{n+1}(1+2\varepsilon)^{-n-1}$ from $A_n$. We show first that this operation yields a well-defined set for each $n$ by showing that the following formula holds:
$$m(A_n)=1-\sum_{i=1}^n2^{i-1}\frac{\varepsilon^i}{(1+2\varepsilon)^i},$$
which is nonnegative for each $n\in\mathbb{N}$, since:
$$1-\sum_{i=1}^n2^{i-1}\frac{\varepsilon^i}{(1+2\varepsilon)^i}\geq1-\sum_{i=1}^n2^{i-1}3^{-i}=1-1=0.$$
The formula holds for $A_1$, as we have:
$$m(A_1)=1-\frac{\varepsilon}{1+2\varepsilon}.$$
If the formula holds for $n=k$, then:
\begin{align*}
m(A_{n+1})&=\left(1-\sum_{i=1}^n2^{i-1}\frac{\varepsilon^i}{(1+2\varepsilon)^i}\right)-2^k\frac{\varepsilon^{k+1}}{(1+2\varepsilon)^{k+1}}\\
&=1-\sum_{i=1}^{k+1}2^{i-1}\frac{\varepsilon^i}{(1+2\varepsilon)^i},
\end{align*}
so the desired formula holds by induction.
Let $A$ be the intersection of all $A_n$. Then $A$ has measure:
$$m(A)=m\left(\bigcap_{n=1}^\infty A_n\right)=1-\sum_{i=1}^\infty 2^{i-1}\frac{\varepsilon^i}{(1+2\varepsilon)^i}=1-\varepsilon.$$
Further, $A$ is closed and is nowhere dense, since the interior of $A$ contains no open interval. Therefore, defining $E=A^c$ yields an open set dense in $[0,1]$ with measure $m(E)=\varepsilon$. $\,\blacksquare$
Is this proof correct?
Proof that $A$ contains no open interval: Suppose there exists an open interval $(a,b)\subseteq A$ where $a<b$. Then $m\big((a,b)\big)=b-a$, so $(a,b)$ has nonzero measure. Take $n$ large enough such that $A_n$ contains no connected component of measure greater than or equal to $b-a$, which is possible since the $2^{n}$ connected components of $A_n$ each have measure $2^{-n}m(A_n)$, and $2^{-n}m(A_n)\to0$ as $n\to\infty$. Then $(a,b)\not\subseteq A_n$, contradicting the assumption that $(a,b)\subseteq A$.
Best Answer
This is not an "answer", just another way to get the set $E$ and a bit long for a comment. If the requirement was $m(E)\le\epsilon$ this is well known: enumerate the rationals in $(0,1)$ as $\{q_n\}_{n\in\mathbb N}$, define $A_t$ ($t\gt 0$) to be the union of the open sets $(q_n-t/2^n,q_n+t/2^n)\cap (0,1)$ and take $E = A_{\epsilon}$.
To get equality, notice that the function $f$ defined by $f(x) = m(A_x)$ is continuous and satisfies $f(2) = 1$, $f(\epsilon/2)\lt\epsilon$, so $f(t) =\epsilon$ for some $t\in(\epsilon/2,2)$.