Clarification on a proof of the Skorokhod representation theorem

Question

(Skorokhod's representation theorem): Let ${X_1,X_2,\dots}$ be a sequence of real random variables, and $X$ a further random variable. Then ${X_n}$ converges in distribution to ${X}$ if and only if, after extending the probability space model if necessary, one can find copies ${Z_1,Z_2,\dots}$ and ${Z}$ of ${X_1,X_2,\dots}$ and ${X}$ respectively such that ${Z_n}$ converges almost surely to ${Z}$.

Use the unit interval with the Borel measure as the common probability space for the copies, and denote the distribution function of $X$ by $F_X$ (and similarly for $X_n$), it is known that

$\displaystyle Z_n(\omega) := \inf \{y \in {\bf R}: F_{X_n}(y) \geq \omega\}, Z(\omega) := \inf \{y \in {\bf R}: F_{X}(y) \geq \omega\}$

are random variables with distribution functions $F_{X_n}$ and $F_X$ resp. Let $\omega \in (0, 1)$. Standard proof such as Theorem $13.1$ in Probability: A Graduate Course proceeds by finding a continuity point $y$ of $F_X$ with $F_X(y) < \omega \leq F_X(y + \varepsilon)$ to conclude that

$\lim \inf_{n \to \infty} Z_n(\omega) \geq Z(\omega)$.

And similarly a continuity point $x$ of $F_X$ with $F_X(x - \varepsilon) \leq \omega^* < F_X(x)$ to conclude that

$\lim \sup_{n \to \infty} Z_n(\omega^*) \leq Z(\omega^*)$

for some $\omega^* \in (\omega, 1)$. And finally combine the two inequalities to conclude that $Z_n(\omega) \rightarrow Z(\omega)$ for all continuity point $\omega$ of $Z$ (which is a countable set as $Z$ is monotone).

Question: Why do some proofs go the extra length to find a different $\omega^* > \omega$ to bound the limsup? It seems that the same argument works by using the same $\omega$ to conclude that $Z(\omega) \leq \lim \inf_{n \to \infty} Z_n(\omega) \leq \lim \sup_{n \to \infty} Z_n(\omega) \leq Z(\omega)$ for any $\omega \in (0, 1)$.

Answer 1

Indeed $Z(\omega)\leq \liminf_nZ_n(\omega),\omega \in (0,1)$. If you want to work with the same fixed $\omega$ then you can obtain $Z(\omega)\leq \liminf_nZ_n(\omega)\leq \limsup_nZ_n(\omega)\leq Z(\omega^+)$ where $Z(\omega^+)=\lim_{\omega' \downarrow \omega}Z(\omega')$ (recall the generalised inverse of a cdf is left-cont. and nondecreasing, so the right-limit exists).

To prove $\limsup_nZ_n(\omega)\leq Z(\omega^+)$: let $\varepsilon>0$. Suppose $\limsup_nZ_n(\omega)> Z(\omega+\varepsilon)$. Then there exists $\eta>0$ such that for all $N \in \mathbb{N}$ there exists an $n\geq N$ so that we have $Z_n(\omega)\geq Z(\omega+\varepsilon)+\eta$. So choose $x_0\in [Z(\omega+\varepsilon),Z(\omega+\varepsilon)+\eta)$ which is a cont. point of $F_X$ and a subsequence $Z_{n_k}(\omega)\geq Z(\omega+\varepsilon)+\eta$ for all $k$. We thus get $Z_{n_k}(\omega)\geq Z(\omega+\varepsilon)+\eta>x_0\geq Z(\omega+\varepsilon)$. So $F_X(x_0)\geq \omega+\varepsilon$ but also $F_{X_{n_k}}(x_0)<\omega$ for all $k$ which implies $F_X(x_0)=\lim_kF_{X_{n_k}}(x_0)\leq \omega$, a contradiction. Therefore, $\limsup_nZ_n(\omega)\leq Z(\omega+\varepsilon)$ for all $\varepsilon>0$ and by taking the limit $\varepsilon\downarrow 0$ we conclude.

There is an example in which $\omega=1/2$ is not a cont. point of $Z$ and $\limsup_nZ_n(\omega)=Z(\omega^+)>Z(\omega)$. Think of $$F_{X_n}(x)= \begin{cases} 0&x<0\\ (1/2-1/2^n)+x/2^n&0\leq x<1\\ 1&x\geq 1 \end{cases}$$