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Let $D=\{x: [0,1]\rightarrow\mathbb{R}| \text{x is right-continuous with left limits(cadlag in French)}\}$.

Then $D$ is Skorokhod Space with Skorokhod distance given by: for any $x, y\in D$,

$$d(x,y)=\inf_{\lambda\in\Lambda}\{\max\{||\lambda-Id||, ||x\lambda-y||\}\}$$

Here $\Lambda$ implies the set of all continuous and strictly increasing functions $\lambda$ defined in $[0,1]$ such that $\lambda(0)=0$ and $\lambda(1)=1$. $||\cdot||$ is the uniform norm and $x\lambda$ is the composition of $x$ and $\lambda$.

Set

$$\omega_x(\delta)=\sup_{t\in[0,1-\delta]}\omega_x([t,t+\delta]),\ \ \omega_x([t,t+\delta])=\sup_{s,s'\in[t,t+\delta]}|x(s)-x(s')|$$

I want to show $||x-y||\leq d(x,y)+\omega_x(d(x,y))\ \ (\ast)$.

For any $\epsilon>0$, we have $\lambda\in\Lambda$ s.t.

$$||x\lambda-y||<d(x,y)+\epsilon,\ \ ||\lambda-Id||<d(x,y)+\epsilon$$

Then for all $t\in[0,1]$,

$$|x(t)-y(t)|\leq |x(t)-x(\lambda(t))|+|x(\lambda(t))-y(t)|\leq \omega_x(||\lambda-Id||)+||x\lambda-y||\leq \omega_x(d(x,y)+\epsilon)+d(x,y)+\epsilon$$

By passing $\epsilon$ to $0$, we get

$$|x(t)-y(t)|\leq\lim_{\epsilon\rightarrow 0}\omega_x(d(x,y)+\epsilon)+d(x,y)$$

So in order to prove $(\ast)$, it is enough to show $\omega_x(\delta)$ is continuous with respect to $\delta$, does someone have an idea? Thanks a lot!

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