Let $X=\mathbb T^d=\mathbb R^d/\mathbb Z^d$ and let $E$ be a proper open subset of $X$. We say $E$ is *geodesically convex* if for any $x,y\in E$ the **shortest** geodesic connecting $x$ and $y$ lies in $E$. 


**Question.** How large can the Haar/Lebesgue measure of $E$ can be?


For example, is $d=2$, then it seems that this cannot exceed $1/2$. Say, $[0,1)\times [0,s)$ is geodesically convex if and only if $s\leq1/2$. (If $s>1/2$, then $[x,x+\delta]$ is not the shortest geodesic for any $\delta\in(1/2,s)$ and any $x\in(0,1)$.) 


Is it true for any $d\ge2$ that the measure of such an $E$ cannot exceed $1/2$? 



Large geodesically convex subsets of tori
*(This is a new answer; my original answer was completely wrong.)*


Assume $\mathop{\rm vol}E>\tfrac12$.
Then it contains two opposite points say $x$ and $x'=x+(\tfrac12,\tfrac12,\dots,\tfrac12)$.
WLOG we can assume that $x=0$.
Taking minimizing geodesics form $(\tfrac12,\tfrac12,\dots,\tfrac12)$ to $y\approx 0$, we get that all main diagonal of unit cube 
$$\square^n=(0,1)\times(0,1)\times\dots\times(0,1)$$ 
lie in $E$. 
Then apply the following lemma:


**Trivial Lemma.** Let $\square^n$ be open unit cube in $\mathbb R^n$ and $E\subset \square^n$ be a locally convex open set which contains all main diagonals of $\square^n$ then $E=\square^n$.


To prove the lemma, note that local convexity + conectedness in $\mathbb R^n$ implies convexity.


