Let the support $S$ of a distribution function $F$ be $$S = \left\{x: F(x+\epsilon)-F(x-\epsilon) > 0, \forall \epsilon>0\right\}$$

I want to show that this is a closed set. In case anyone needs a reminder, a distribution function is non-decreasing, continuous from the right, limit to infinity is 1, limit to negative infinity is 0.

ATTEMPT: My analysis is a bit rusty. A set is closed if it contains all of its limit points. However, if I take that approach I can't see a way to show that if $y$ is close enough to $x$ then it also has the above property without assuming $F$ is continuous (which it doesnt have to be).

I also know the complement of an open set is closed, so I could also try showing $S^c$ is open.

Any tips or suggestions?


1 Answer 1


Suppose $x\in S^C$. Then there exists $\epsilon$ such that $F(x+\epsilon)-F(x-\epsilon) = 0$, i.e. $F(x+\epsilon)=F(x-\epsilon)$. Since $F$ is non-decreasing, $F$ is constant in $(x-\epsilon,x+\epsilon)$. For $y\in (x-\epsilon,x+\epsilon)$, let $\delta = \min{(|y-(x-\epsilon)|,|(x+\epsilon)-y|)}$ so that $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$. Therefore $F(y+\delta)=F(y-\delta)$ and so $y\in S^C$ which means $S^C$ is an open set, and thus $S$ is a closed set.


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