How to show that in a subset of [0,1] of measure greater than 0.5, there exist two points at distance exactly 0.1? My attempt: Let's disregard isolated points as they do  not contribute to measure. If our set is union (disjoint) of finitely many, say $n$ intervals, there must be at least ($n-1$) intervals of length at least 0.1 not in our set (the 0.1 intervals to the right of the intervals in our set, except the end one possibly). 
Since our set has measure 0.5, 0.1($n-1)<0.5\Rightarrow n\leq 5$, but each such interval is at most 0.1 long. This contradicts that our set has measure 0.5. If the set is a union of countably infinite many intervals, the situation is worse. (how exactly do I argue this?) 
I think uncountably many intervals is impossible as no positive measure subset of [0,1] is a union of uncountably many positive measure sets. Am I correct in my approach? What else is needed to complete the solution? Thanks for any help.
 A: You divide $[0,1]$ as : $[0,\frac{1}{10}[ \cup [\frac{1}{10}, \frac{2}{10}[\cup\cdot \cdot \cdot \cup [\frac{9}{10}, 1]$
Noting $A$ your set, you now observe what are the measures of the intersections of A and any of these intervals of length $\frac{1}{10}$.
Let us denote $\mu_1$ the measure of $A \cap [0,\frac{1}{10}[$, ..., $\mu_{10}$ the measure of $A \cap [\frac{9}{10}, 1]$. 
Now we assume the contrary of what you wishes to prove. Using that $\mu$ is translation-invariant, you get that $\mu_2 \leqslant \frac{1}{10} - \mu_1$, because $A\cap [\frac{1}{10},\frac{2}{10}[ \subset [\frac{1}{10},\frac{2}{10}[ \backslash (A\cap[0,\frac{1}{10}[ + \frac{1}{10})$ : if you add $0.1$ to any element of the first set, it will not be in $A$.
So $\mu_1 + \mu_2 \leqslant \frac{1}{10}$. More generally, for $1 \leqslant i \leqslant 9$, $\mu_i+\mu_{i+1} \leqslant \frac{1}{10}$.
Using this inequality for $i=1,3,5,7,9$ and summing, we have : $\sum \limits_{i=1}^{10} \mu_i \leqslant 5\cdot \frac{1}{10}$.
As the sum of thoses measures of distinct sets is precisely equal to $\mu(A)$, we deduce $\mu(A) \leqslant \frac{1}{2}$.
So finally if $\mu(A)>0.5$, the result is true.
A: A measurable subset $A\subset[0,1]$ is not necessarily a disjoint union of isolated points and intervals, it may have the structure of a (fat) Cantor set. Anyway, if we assume that $A$ and $\left(A+\frac{1}{10}\right)\cap[0,1]$ are disjoint subsets of $[0,1]$, the sum of their measures has to be $\leq 1$.
If we know that $\mu(A)>\frac{1}{2}+\frac{1}{20}$, since $\mu$ is translation-invariant, we have $A\cap\left(A+\frac{1}{10}\right)\neq\emptyset$.
The above argument has to be adjusted if we just know that $\mu(A)>\frac{1}{2}$. 
In such a case, for $i\in\{1,2,3,4,5\}$, let $A_i=\left(\frac{i-5}{5},\frac{i}{5}\right)$ and 
$$\delta_i = \frac{\mu(A\cap A_i)}{\mu(A_i)}. $$
Since $\mu(A)>\frac{1}{2}$, at least one $\delta_i$ (relative density) is greater than $\frac{1}{2}$. We may assume WLOG that $\delta_1>\frac{1}{2}$ and split $A_1$ in two halves, $L=\left(0,\frac{1}{10}\right)$ and $R=\left(\frac{1}{10},\frac{2}{10}\right)$, then define:
$$\delta_L = \frac{\mu(A\cap L)}{\mu(L)},\qquad \delta_R = \frac{\mu(A\cap R)}{\mu(R)}.$$
We have $\delta_L+\delta_R>1$, hence $L+\frac{1}{10}$ and $R$ cannot be disjoint.
A: Consider $A\cap\{[0,0.1)\cup[0.2,0.3)\cup[0.4,0.5)...\}$
