# proving a function satisfies a Lipschitz condition

Let $F$ be a closed set of $\mathbb{R}$ whose complement has finite measure. Let $\delta(x) = d (x, F) =\inf \{ |x - z| \mid z \in F\}$.

Prove $\delta$ continuous by proving $| \delta(x) - \delta(y) | \leq |x - y|$

I appreciate any kind of hint. Thanks

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Let $x, y\in\mathbb{R}$. For any $\epsilon>0$, by definition of $\delta(y)$, there exists $z\in F$ such that $$|y-z|\leq\delta(y)+\epsilon.$$ As Ashok said, we have the triangle inequality $$|x-z|\le |x-y|+|y-z|.$$ Combining these two inequalities, we have $$|x-z|\le |x-y|+\delta(y)+\epsilon.$$ By definiton of $\delta(x)$, we have $\delta(x)\leq |x-z|$ since $z\in F$. Hence, we have $$\delta(x)\le |x-y|+\delta(y)+\epsilon.$$ Since $\epsilon>0$ is arbitrary, we have $\delta(x)\le |x-y|+\delta(y)$, or $$\delta(x)-\delta(y)\le |x-y|.$$ By symmetry, we also have $$\delta(y)-\delta(x)\le |y-x|.$$ Therefore, we obtain $$| \delta(x) - \delta(y) | \leq |x - y|.$$
I don't think we need $F$ to be a closed set whose complement has finite measure etc. The result holds for any nonempty $F$.
Hint: $|x-z|\le |x-y|+|y-z|$ and take $\inf$ over $z\in F$.