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Let $(M,\rho)$ be a metric space, $F\subset M$ a closed subset, $\varepsilon>0$.

On page 8 of the book "Convergence of Probability measures" by Patrick Billingsley one says that the function $$f:M\to [0,1]\\ x\mapsto(1-\rho(x,F)/\varepsilon)^+$$ is uniformly continuous since it holds $|f(x)-f(y)|\leq \rho(x,y)/\varepsilon$.

Actually I'm not able to prove this inequality: I tried to solve all the possible cases ($x,y\in F, \not\in F, \in F^\varepsilon, \not\in F^\varepsilon$) but it doesn't looks elegant at all and I even could't prove all the possibilities.

$(.)^+:=max(0,.)$

Thanks

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    It looks like a triangle inequality application. Perhaps it would help if you identified one of the cases you solved as well as one you could not. – 2017-01-14

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If $x$ is in the metric space and $z \in \mathrm{F}$ then $d(x, \mathrm{F}) \leq d(x, z) \leq d(x, y) + d(y ,z)$ so taking the infimum on $z$ on the left-most and right-most gives $d(x, \mathrm{F}) - d(y, \mathrm{F}) \leq d(x,y)$ and by symmetry, you can put absolute value. So $f(x) - f(y) = \dfrac{\rho(x, \mathrm{F}) - \rho(y, \mathrm{F})}{\varepsilon}.$ Is the rest clear from here?

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    The rest is clear, thank you! – 2017-01-14