Let $\epsilon>0$.Construct an open set $E$ $\subseteq$ $\mathbb{R}$ such that $m(E\cap I)>0$ for every I interval in $\mathbb{R}$ and $m(E)\leqslant\epsilon$, where the symbol $m$ denotes the lebesgue measure
What i did:
Let $\epsilon>0$ and $\mathbb{Q}=\{q_1,q_2,,,q_n,,,\}$.
I took $E=\bigcup_{n=1}^{\infty}(q_n-\epsilon/2^{n+1},q_n+\epsilon/2^{n+1})$.
By doing some calculations and using the lebesgue measure's subadditivity, we can easily see that $m(E)\leqslant \epsilon$
Let $I=(a,b)$ ,$a<b$.From the density of the rationals in the real line $\exists q\in \mathbb{Q}$ such that $a<q<b$.
Also $\exists n \in \mathbb{N}$ such that $q=q_n$.
Now we notice that $a-\epsilon/2^{n+1}<q_n-\epsilon/2^{n+1}<q_n+\epsilon/2^{n+1}<b+\epsilon/2^{n+1}$.
If we take $E\cap I=(c,d)$ were $c=max\{q_n-\epsilon/2^{n+1},a\}$ and $d=min\{q_n+\epsilon/2^{n+1},b\}$ then we have $m(E\cap I)>0$.
For the case where $I=(-\infty,a)$ and $J=(b,+\infty)$ the we can express $I=\bigcup_{n=1}^{\infty}(-n,a)$ and $J=\bigcup_{n=1}^{\infty}(b,n)$ and apply the above argument.
The same arguemt applies for closed intervals.
Is this proof correct or am i missing something here?
Can someone help?
Thank you in advance!