In the book Probability: Theory and Examples by Durrett, on page 118 he claims the following (I have adapted the notation and verbiage slightly for simplicity and clarity):
If distribution functions $F_n$ converge weakly to a distribution $F$, then there exist random variables $Y$ and $Y_n$ having $F$ and $F_n$ as their respective distribution functions, and such that $Y_n \rightarrow Y$ almost surely.
I think there is an error in the proof (which I have reproduced below), but it is late in the day and perhaps I am just missing something. My question is: Is the critique below of Durrett's proof correct? If so, how can we patch up this proof?
Proof: Set $\Omega = (0,1)$, $\mathcal{F}=$ Borel sets, $P=$ Lebesgue measure, and $Y_n(x) = \sup\{y : F_n(y)<x\}$ (and similarly for $Y$). By another result in Durrett, $Y_n$ so defined has $F_n$ as its distribution function. We will show that $Y_n\rightarrow Y$ for all but a countable number of points $x$.
For all $x\in \Omega$ set $a_x = \sup\{y : F(y)<x\} = Y(x)$ and $b_x = \inf\{y : F(y) > x\}$. Let $\Omega_0 = \{x : (a_x,b_x)=\emptyset \iff a_x = b_x\}$. Then $\Omega-\Omega_0$ is countable, because the non-null intervals $(a_x, b_x)$ are disjoint and open (and thus each contain a distinct rational point, of which there are countably many). Hence $\Omega_0$ has full measure, and we claim that $\forall x\in\Omega_0$ we have $Y_n(x)\rightarrow Y(x)$.
To prove convergence, we will show that $\forall x\in \Omega_0$ we have $\liminf_{n\rightarrow \infty} Y_n(x) \geq Y(x)$ and $\limsup_{n\rightarrow \infty} Y_n(x) \leq Y(x)$. To prove the former, let $z<Y(x)$ be an arbitrary continuity point of $F$. From the definition of $Y$ as the sup of points $y$ s.t. $F(y)<x$, we see that $F(z)<x$. Hence for $n$ sufficiently large $F_n(z)<x$ (by weak convergence), and thus $Y_n(x)\geq z$. This holds for all such $z<Y(x)$, and thus taking the limit (using the fact that $F$ has only countably many discontinuities) we see that $\liminf_{n\rightarrow \infty} Y_n(x)\geq Y(x)$ as desired.
Now to show that $\limsup_{n\rightarrow \infty} Y_n(x) \leq Y(x)$, we similarly pick an arbitrary $z > Y(x)$ which is a continuity point of $F$. Durrett then claims (and this is the point I dispute, for reasons described below) that we must have $F(z) > x$. If this holds, then we can argue as in the previous case: By weak convergence of $F_n$, we have that $F_n(z) > x$ for large $n$, hence $Y_n(x) = \sup\{ y : F(y) < x\} \leq z$, hence $\limsup_{n\rightarrow \infty} Y_n(x) \leq Y(x)$ as desired.
Now my beef with the disputed claim is this: If $F$ is flat over some interval $[a,b)$ then for $x=F(a)$ and $z\in (a,b)$ we have $z>Y(x)=a$, but do not have $F(z) > x$, for in fact $F(x) = x$. As a concrete counter-example, consider $F = \frac{1}{2}\left(1_{[0,\infty)} + 1_{[1,\infty)}\right)$. Then $Y(x) = 0$ for $x\in \left(0,\frac{1}{2}\right]$ and $Y(x) = 1$ for $x\in \left(\frac{1}{2},1\right)$. So for $x=F(0) = \frac{1}{2}$ we have $Y(x) = 0$, and for $z=\frac{1}{2} > Y(x)$ we have $F(z) = \frac{1}{2} = x$.
