Let $\lambda:=\lambda^1_{|[0,1]}$ be Lebesgue measure on ([0,1],$\mathcal{B}$[0,1]). Show that for every $\textbf{$\epsilon$}>0$ there is a dense open set $ U $ with $\lambda(U)\leq\textbf{$\epsilon$}$.

[Hint: take an enumeration $(q_j)_{j\in \mathbb{N}}$ of $\mathbb{Q}\cap(0,1)$ and make each $q_j$ the centre of a small open interval.]

Following the hint I can take the union of intervalls around each $q_j$ but that would make infinitely many sets all with finite measure. So how would one prove this question?


Following @aduh's hint: Let $\{q_n\}$ be a, necessarily countable, enumeration of the rational numbers in $(0,1)$. For every $n$, take the open set $A_n = (q_n - \varepsilon/2^n, q_n + \varepsilon/2^n)$. Then we have that $\lambda(A_n) = \varepsilon/2^n$

Then, letting $A=\cup_n A_n$ we have $$\lambda(A) = \lambda\left( \bigcup_{n=1}^{\infty} A_n\right) \leq \sum_{n=1}^{\infty} \lambda (A_n) = \sum_{n=1}^{\infty} \varepsilon/2^n = \varepsilon$$

As required.


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