Set of first category in $\mathbb{R}$ such that the intersection with an open set has positive Lebesgue measure

Question

There exists and $F_\sigma$-set $B\subset \mathbb{R}$ that is of first category in $\mathbb{R}$ which has the further property that $\lambda(U\cap B)>0$ and $\lambda(U\cap B')>0$ for every nonvoid open subset $U$ of $\mathbb{R}$. ($\lambda$ is the Lebesgue measure on $\mathbb{R}$.)

I have some questions on the proof of this theorem, specifically in the construction to find such a set. One starts by enumerating all the nonvoid open intervals $\lbrace I_n \rbrace_{n=1}^\infty$ having rational endpoints. First we choose $s_1=\lambda(I_1)$ and $B_0=\emptyset$. Inductively, when a compact nowhere dense set $B_{n-1}$ and positive numbers $s_1,\dots,s_n$ have been selected ($n$ is a natural number), we choose a compact nowhere dense set $A_n\subseteq I_n\cap B'_{n-1}$ with $0<\lambda(A_n)<2^{-n}\min \lbrace s_1,\dots,s_n\rbrace$. Write $B_n = B_{n-1}\cup A_n$ and $s_{n+1}=\lambda(I_{n+1})-\lambda(I_{n+1} \cap B_n )$. (proof continues)

So far I think I'm missing the big picture. Even if $\lambda(A_n)<2^{-n}({\rm etc})$, it seemes to me that the $A_n$ can have a rather wild measure. I guess they are chosen such that the measure goes to zero, but I cannot see this. Connected to this, I don't really see the motivation to choose $s_{n+1}$ they way it is, it might be $I_{n+1}\cap B_n=\emptyset$ but $\lambda(I_{n+1})$ can still be quite big.

Answer 1

There are cantor sets (nowere dense sets) that have positive measure (i can provide an example if you want), so take that set $C$ in $[0,1]$ such that $\lambda(C)=a>0$, also take $C/n$ (where $A/n=\lbrace a/n:a\in A\rbrace$) that has measure $a/n$, now define the set $C_n=\bigcup_{i=0}^{n-1}C/n+i/n$ this is a set contained in $[0,1]$, now take the set $K_0=\bigcup C_n\subset [0,1]$, since $K$ is a countable union of nowhere dense sets it is of first category, and $K_n=K_0+n$, $K=\bigcup_{n\in\mathbb Z} K_n$, since countable union of first category sets are of first category you have that $K$ is of first category. Now take an interval $(a,b)$, then there exists a rational $p/q\in [0,1)$, and $n\in\mathbb Z$ such that $\left[n+\frac{p}{q},n+\frac{p+1}{q}\right]\subset (a,b)$, then $C_q+p/q+n$ is contained in $(a,b)$, since $C_q+p/q+n\subset K$ and $\lambda(C_q+p/q+n)=a/q>0$ you have your proof.