I am working on a slight variation of the proof of the Skorokhod representation theorem, as found in Bogachev, Vol. II. In particular, I want to use another definition of the functions $F_\mu,\xi_\mu$.
Let $X=[0,1]$, $\mu$ a probability measure on $(X,\mathscr{B}(X))$; define $F_\mu(x):=\mu([0,x])$ and $$\xi_\mu(t)=\inf\{x \in [0,1]:F_\mu(x)\geq t\}$$ Let the sequence of probability measures $\mu_n\to \mu$ weakly. I would want to prove $\xi_{\mu_n}\to \xi_\mu$, $\lambda$-a.e. where $\lambda$ is the Lebesgue measure. I start with the following facts:
- $\xi_\mu(t)\leq x\iff F_\mu(x)\geq t$ , which implies $\xi_\mu(t)> x\iff F_\mu(x)< t$;
- $\xi_\mu$ is left-continuous nondecreasing: the set of discontinuity points is countable.
- $F_{\mu_n}(x)\to F_\mu(x)$ at all continuity points of $F_\mu$; again the set of discontinuity points is countable.
Now I proceed: what could be the issues with this argument?
Suppose $$\liminf_{n \to \infty}\xi_{\mu_n}(t)<\xi_\mu(t)-2\varepsilon$$ We choose $x_0\in (\xi_\mu(t)-2\varepsilon,\xi_\mu(t)-\varepsilon)$ a continuity point of $F_\mu$. Then, for a subsequence, $\xi_{\mu_{n_k}}(t)<x_0,\,\forall k$. Then $t\leq F_{\mu_{n_k}}(x_0)\to F_\mu(x_0)\implies \xi_\mu(t)\leq x_0$, contradiction. So $\liminf_{n \to \infty}\xi_{\mu_n}(t)\geq \xi_\mu(t)$. Suppose that for some $\eta >0$ $$\limsup_{n \to \infty}\xi_{\mu_n}(t)>\xi_\mu(t+\eta)+2\varepsilon$$ We choose $x_0\in (\xi_\mu(t+\eta)+\varepsilon,\xi_\mu(t+\eta)+2\varepsilon)$ a continuity point of $F_\mu$. Then, for a subsequence, $\xi_{\mu_{n_k}}(t)>x_0,\,\forall k$. We get $$\begin{aligned}\xi_{n_k}(t)>x_0&\implies F_{\mu_{n_k}}(x_0)<t\implies \\ &\implies F_\mu(x_0)\leq t<t+\eta\implies \\ &\implies \xi_{\mu}(t+\eta)> x_0\end{aligned}$$ a contradiction. We obtain $$\xi_\mu(t)\leq \liminf_{n \to \infty}\xi_{\mu_n}(t)\leq \limsup_{n \to \infty}\xi_{\mu_n}(t)\leq\xi_\mu(t^+) $$ So if $t$ is a continuity point of $\xi_\mu$, we have $\xi_{\mu_n}(t)\to \xi_\mu(t)$.