Proof of Theorem 13.3 in Convergence of Probability Measures by Billingsley I have some trouble understanding the "symmetric argument" part in the proof of Theorem 13.3. (how to get the 3rd condition in Equation 13.8 from Equation 13.9). According to the book, some symbles are defined as:
$D$: the space of cadlag functions: real valued functions on $[0,1]$ which are right continuous and have left limit at $t\in(0,1]$.
$P_n,P$: probability measures on the space $D$.
$T_P$: The set of $t\in [0,1]$ for which the projection $\pi_t$ is continuous with respect to the Skorohod topology except at points forming a set of P-measure 0 ($T_P$ is dense in [0,1] and contains 0 and 1).
$J_1$: the set of $x \in D$ where $x(1-)\neq x(1)$.
$\omega_x^{''}(\delta)=\sup_{^{t_1\leq t \leq t_2}_{t_2-t_1\leq\delta}}\,\,\,\{|x(t_1)-x(t)|\wedge|x(t)-x(t_2)|\}$
Some equations and the proof are in the following pictures. The main part of the proof is to show that 2nd and 3rd condition in (13.8) holds for large n.
I don't know if it is true to say that by proving $P(J_1)=0$, the author is saying that somehow $1-$ can be viewed as a point in $T_P$ and $P_n\pi_{1-}^{-1} \Rightarrow P\pi_{1-}^{-1}$?
In the discussion following Theorem 13.3, the author says 13.9 can be replaced by $P_n\pi_{1-}^{-1} \Rightarrow P\pi_{1-}^{-1}$.

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 A: I think what he means by "symmetric argument" here is that in order to show that for $\epsilon,\eta > 0$, there exists $n_0, \delta$ so that for $n\ge n_0$,   $P_n[x: |x(1-)-x(1-\delta)| \ge \epsilon ] < \eta$, the following is established:

*

*$P_n \pi_{\delta,1}^{-1} \to  P \pi_{\delta,1}^{-1}$, $\delta \in T_P$. This is assumed to hold.


*$P[x: |x(-1)-x(1-\delta)|\ge \epsilon ] \to 0$ as $\delta \to 0$. He gives the argument for this.


*$P[x: |x(1)-x(1-)|\ge \epsilon ] = 0$. He gives the argument for this.
These can be combined to show item 3) of 13.8. The argument is "symmetric" in the sense that the first argument used $P_n \pi_{0,\delta}^{-1} \to  P \pi_{0,\delta}^{-1}$, the second half uses $P_n \pi_{\delta,1}^{-1} \to  P \pi_{\delta,1}^{-1}$, but there is the added complication in this case that functions in $D$ only have left limits, so dealing with $x(1-)$ is a bit more delicate/tedious.
A: I kind of figured out the answer, but I'm not sure. (13.9) alone will not work, must combine it with (13.10):
For any $\varepsilon>0$
\begin{align}
&P_n(|x(1-)-x(1-\delta)|\geq 2\varepsilon)\\
=&P_n(|x(1-)-x(1-\delta)|\geq 2\varepsilon;|x(1-)-x(1)|< \varepsilon)+P_n(|x(1-)-x(1-\delta)|\geq 2\varepsilon;|x(1-)-x(1)|\geq \varepsilon)\\
\leq &P_n(|x(1-\delta)-x(1)|\geq \varepsilon)+P_n(|x(1-)-x(1-\delta)|\wedge|x(1-)-x(1)|\geq \varepsilon) \\
\leq& P_n(|x(1-\delta)-x(1)|\geq \varepsilon)+ P_n(\omega_x^{''}(\delta)\geq\varepsilon)
\end{align}
if we choose $\delta$ small enough and $1-\delta \in T_P$, then the 1st term in the last inequality can be bound by (13.9) and weak convergence of $P_{n\pi^{-1}_{1-\delta,1}} \Rightarrow P_{\pi^{-1}_{1-\delta,1}}$, and the 2nd term in the last inequality can be bound by (13.10)
Here is an example of why (13.10) is needed. Define
$$x_n(t) =
\begin{cases}
0,  & t\in[0,1-\frac 1n] \\
\text{linear from 0 to 1}, & t\in (1-\frac 1n,1)\\
0, &t=1
\end{cases}$$
and $x(t)=0, t\in[0,1]$. Define $P_n$ and $P$ to be the point mass on $x_n$ and $x$. Then (13.9), as well as the weak convergence of $P_n\pi_t^{-1}$for $t\in T_P$ is satisfied. but 3) of (13.8) doesn't hold for $\varepsilon<1$.
