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Let $A \subset [a,b]$ with $m(A)>\dfrac 12 (b-a)$ , where $m$ is the Lebesgue measure . Then is it true that $A$ contains a subset of positive measure which is symmetric about $(a+b)/2$ ?

( we say that a subset $S$ of real line is symmetric about a real number $c$ if $c+x \in S \iff c-x \in S$ )

My idea was like to take $T:=\{x \in [a,b] : x+(a+b)/2 , (a+b)/2-x \in A\}$ and then take $B:=(\dfrac {a+b}2 +T) \cup (\dfrac {a+b}2 -T)$ ; definitely $B$ is the largest symmetric set about $(a+b)/2$ in $A$ . But I don't know how to get the required set from here . Please help . Thanks in advance

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To keep thinks simple, consider the interval $[-1/2,1/2].$ Suppose $A$ is a measurable subset of $[-1/2,1/2]$ with $m(A)> 1/2.$ Then $-A$ is a measurable subset of $[-1/2,1/2],$ and $m(-A) = m(A).$

Let $B = A\cap (-A).$ If we show $m(B) > 0,$ we're done, since $B$ is symmetric about $0.$

Suppose to the contrary that $m(B)=0.$ Then $A\setminus B$ and $(-A)\setminus B$ are disjoint and measurable, and each has measure $m(A).$ Thus

$$1= m([-1/2,1/2]) = m([-1/2,1/2]\setminus B) \ge m(A\setminus B)+ m((-A)\setminus B)$$ $$= m(A) + m(-A) > 1/2+1/2 =1,$$

contradiction.

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    we could have also finished like this : since $A \cup (-A) \subseteq [-1/2,1/2]$ , so $m(A \cup (-A)) \le 1$ hence $m(A \cap (-A)) =m(A)+m((-A))-m(A \cup (-A))2m(A)-m(A \cup (-A))>2.1/2-1=0$ – 2017-01-31