# A proof that Skorohod metric is a metric

I'm reading Billingsley's "Convergence of probability measures" (1968), p. 111. The definitions are: $D$ - the space of $\textit{cadlag}$ functions on [0,1], $\Lambda$ - the class of strictly increasing, continuous mappings of [0,1] onto itself. For $x,y\in D$ define $$d(x,y):=\inf\{\varepsilon>0:\ \exists\lambda\in\Lambda:\ \sup_t|\lambda(t)-t|<\varepsilon \text{ and } \\ \sup_t|x(t)-y(\lambda(t))|<\varepsilon\}.\tag{1}$$ I'm stuck with the proof that $d(x,y)=0$ implies $x=y$. The author claims: "$d(x,y)=0$ implies that for each $t$ either $x(t)=y(t)$ or $x(t)=y(t-)$, which in turn implies $x=y$."

This is what I tried: let $x,y\in D$, $d(x,y)=0$ and $t\in[0,1]$. Now from (1) it follows that there exists a sequence $(z_n)\subset[0,1]$ such that $z_n\to t$ and $y(z_n)\to x(t)$ as $n\to\infty$. If $t$ is a continuity point of $y$, then $y(z_n)\to y(t)$, thus $x(t)=y(t)$. If $y$ has a jump at $t$ and $(z_n)$ has a subsequence $(z_{n_k})$ such that $z_{n_k}\geq t, \forall k$, then $y(z_{n_k})\to x(t)$ and $y(z_{n_k})\to y(t)$, as $k\to\infty$, thus $y(t)=x(t)$. Otherwise, $z_n<t$ for $n$ large enough and $y(z_n)\to y(t-)$, as $n\to\infty$, thus $x(t)=y(t-)$.

Now suppose that $y=\mathbf{1}_{[t,1]}$, then $y$ has a jump at $t$ and if $d(x,y)=0$, then $x=\mathbf{1}_{(t,1]}$. But this is a contradiction since $x\neq y$ and $x$ is not $\textit{cadlag}$.

Where am I wrong?

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Why do you say that if $d(x,y)=0$ then $x=\mathbf 1_{(t,1]}$? Your proof shows that $x(s)=y(s)$ for every $s\ne t$. Since $x$ is càdlàg, this implies that $x(t)=\lim\limits_{s\to t^+}x(s)=1=y(t)$. Hence $x=y$. – Did Jun 27 '12 at 15:24
According to the proof given, $x(t)=y(t-)$, since $y$ has a jump at $t$. Thus $x(t)=0$. – nokiddn Jun 27 '12 at 16:28
No, the proof says that, when $y$ has a jump at $t$, either $x(t)=y(t)$ or $x(t)=y(t-)$. Since $x$ is càdlàg, $x(t)=y(t)$. – Did Jun 27 '12 at 22:26
Thank you, I get it now. Will you post your comment as an answer for me to accept, or should I delete the question? – nokiddn Jun 28 '12 at 6:33

One cannot say that if $d(x,y)=0$ then $x=\mathbf 1_{(t,1]}$. Your proof shows that $x(s)=y(s)$ for every $s\ne t$. Since $x$ is càdlàg, this implies that $x(t)=\lim\limits_{s\to t^+}x(s)=1=y(t)$. Hence $x=y$.

More precisely, the proof says that, when $y$ has a jump at $t$, either $x(t)=y(t)$ or $x(t)=y(t-)$. Since $x$ is càdlàg, $x(t)=y(t)$.

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