# Durrett Theorem 3.2.8 (Are inverse distribution functions right continuous?)

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 $$Y_n(x) = \sup\{y : F_n(y) (and similarly for $$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 $$a_x = \sup\{y : F(y) and $$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 $$z be an arbitrary continuity point of $$F$$. From the definition of $$Y$$ as the sup of points $$y$$ s.t. $$F(y), we see that $$F(z). Hence for $$n$$ sufficiently large $$F_n(z) (by weak convergence), and thus $$Y_n(x)\geq z$$. This holds for all such $$z, and thus taking the limit (using the fact that $$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 $$F(z) = \frac{1}{2} = x$$.

The essence of the issue is that, while the disputed claim is not true in general, it is true on the set $$\Omega_0$$. Indeed, that is the only reason we restrict to $$\Omega_0$$ in the proof.
Here's a proof that $$\forall (x\in \Omega_0, z > Y(x))$$ we have $$F(z) > x$$: By the construction of $$\Omega_0$$ we have that $$\inf\{ y : F(y) > x\} = b_x = a_x = Y(x)$$. Since $$z > Y(x) = \inf\{y:F(y) > x\}$$ and $$F$$ is non-decreasing, we must have $$F(z) > x$$, as desired.