1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.