Skorokhod convergence VS uniform convergence Skorohod convergence does not imply uniform convergence.
Billingsley quotes a counterexample: for $0\leq\alpha<1$ the sequence
$x_n(t)=1_{[0,\alpha +\frac{1}{n})}(t)$ does not converge uniformly to $x(t)=1_{[0,\alpha)}(t)$, but it does so in the Skorokhod topology.
I am new to the Skorohod topology. I am not able to find a suitable sequence of increasing continuous functions $\lambda_n$ from $[0,1]$ onto itself converging to the identity. $\lambda_n(t)=t+\frac1n$ was my original guess, but I cannot choose it because it does not map $[0,1]$ onto itself.
Which sequence should one choose here?
 A: In principle your idea is correct, but as you already mentioned the function is not bijective. So, basically, we are looking for a monotone bijective function $\lambda_n:[0,1] \to [0,1]$ such that $$\sup_{t \in [0,1]} |\lambda_n(t)-t| \qquad \text{and} \qquad \sup_{t \in [0,1]} |x_n \circ \lambda_n(t)-x(t)|$$ converge to $0$ as $n \to \infty$. 

Lemma: There exists a bijective mapping $\lambda_n:[0,1] \to [0,1]$ such that $$\sup_{t \in [0,1]} |x_n \circ \lambda_n(t)-x(t)| =0 \tag{1}$$ and $$\sup_{t \in [0,1]} |\lambda_n(t)-t| \to 0. \tag{2}$$

Proof: First of all, note that $(1)$ holds if $\lambda_n$ satisfies $\lambda_n(\alpha) = \alpha+1/n$. We define $\lambda_n$ on the interval $[0,\alpha]$ by
$$\lambda_n(t) := \left(1+ \frac{1}{n \alpha} \right) \cdot t =: c_n \cdot t.$$
Then $\lambda_n:[0,\alpha] \to [0,\alpha+1/n]$ is a monotone bijective mapping satisfying $\lambda_n(\alpha)=\alpha+1/n$. It remains to find a bijective continuation on the interval $[0,1]$. We simply use a linear interpolation between the points $(\alpha,\alpha+1/n)$ and $(1,1)$; that is
$$\lambda_n(t) = \left(\alpha+ \frac{1}{n} \right) + \frac{1-\alpha-1/n}{1-\alpha} (t-\alpha)$$
for $t \in [\alpha,1]$. By virtue of our choice, $\lambda_n$ is an increasing bijective function. Here is how the function $\lambda_n$ looks like:
$\hskip1.5in$ 
If we let $n \to \infty$, then $c_n \to 1$ and therefore it is not difficult to see that $(2)$ holds.
