In the third page of My Scottish Book 'Problems', Paul Erdös writes that
There is a very simple theorem of Steinhaus which says that the difference set of a set of positive measure (say on the line) contains an interval. ... It follows instantly from the Lebesgue density theorem, and therefore by this method one obtains the following theorem: Any set of positive measure on the line, or, more generally, in k-dimensional space, contains all finite sets in the space to within a similarity transformation. The proof is almost immediate because by the Lebesgue density theorem there is an interval or a sphere in which the density is as close to 1 as one wishes, and therefore it follows that set will contain a set which is similar to any finite set.
I just cannot follow the reasoning here-- why does that theorem follows from Lebesgue density theorem?
Any hint or reference will be appreciated!
Let $B$ be a set of positive measure, and by translation and scaling we can assume that $0\in B$ is a point of density $1$ in $B$, that $x_1=0<x_2<...<x_n=1$ is the finite set, and we can also assume that $1\in B$. Scaling furter we can assume that $B\cap [0,1]$ has measure $>1-\epsilon$, that $B\cap[0,x_{n-1}]$ has measure $>x_{n-1}-\epsilon$, and so on, $B\cap[0,x_2]$ has measure $> x_2-\epsilon$.
Assume that for all $r\in[0,1]$, there is an $i_r\in\{1,2,...,n\}$ such that $rx_i\notin B$.
The points of the form $rx_n\notin B$ can only be a set of measure $\epsilon$. The set of points of the form $rx_{n-1}\notin B$ can only be a set of measure $\epsilon$, ... and likewise for all the others. Therefore, the set of values of $r$ for which at least one $rx_i\notin B$ can at most have measure $n\epsilon$. Choosing $\epsilon << 1/n$, we get that there must be some $r\in[0,1]$ such that all $rx_1,...,rx_n\in B$.