Why can we find a Cramér-von Mises test of level $\alpha$? I'm reading through Shao's Mathematical Statitics section 6.5.2. on Cramér-von Mises Tests and there is something I don't understand.
Context :
Let $X_1,...,X_n$ be iid from a continuous cdf $F$. We wish to test if the sample comes from some continuous cdf $F_0$:
$$ 
\begin{cases}
H_0 : F = F_0 \\
H_1 : F \neq F_0. 
\end{cases}
$$
Cramer-von Mises tests are tests which reject $H_0$ when $C_n(F_0) > c$ where
$$ C_n(F) = \int (F_n(x)-F(x))^2 \; dF(x).$$
Question :
On page $448$ we read the following " ... the distribution of $C_n(F)$ does not depend on $F$ (exercice). Hence a Cramer-von Mises test of size $\alpha$ can be found. "
I've managed to solve the exercice by showing that
$$ C_n(F) =  \frac{1}{12n^2} + \frac 1 n \sum_{i=1}^{n}\left(U_i-\frac{2i-1}{2n}\right)^2$$
where $U_i = F(X_{(i)})$ is uniformaly distributed on $[0,1].$ However I don't understand why this implies that a test of level $\alpha$ can be found. Why is this the case ?
If I'm not mistaken this means we can find $c$ such that
$$P_{H_0}(RH_0) = P(C_n(F_0) > c ) = \alpha.$$
How do we know such a $c$ exists ?
 A: If I'm not wrong, this goes back to the idea of a pivot (or a pivotal quantity).
When you want to construct a statistical test it is necessary to construct a confidence interval and pivots are the theoretical way to do this.
A pivot is a function of a sample that does not depend on the underlying distribution function.
For example let $X_1,\ldots,X_n$ be iid with common cdf $F_X$.
Then a pivot $R = R(X_1,\ldots,X_n;F_X)$ is such that
\begin{align}
F_R(x) = \mathbb{P}(R(X_1,\ldots,X_n;F_X) \leq x)
\end{align}
does not depend on $F_X$.
The best example for a pivot is
$$R = \sqrt{n}\frac{\bar{X}_n - \mu}{\sigma}$$
when $X_i \sim N(\mu,\sigma^2)$ and $\bar{X}_n = \frac{1}{n} \sum_{i=1}^n X_i$.
Here we clearly have a dependence on the sample, while $\mu$ and $\sigma$ are functions of $F_X$.
Nonetheless, the distribution of $R$ does not depend on $F_X$ in any way since we have $R \sim N(0,1)$ irrespective of the mean and variance of $F_X$.
In case of a pivot we can find (because $F_R$ is strictly increasing) a constant $c$ independent of $F_X$ such that $1-F_R(c) = \mathbb{P}(R(X_1,\ldots,X_n;F_X) > c) \geq \alpha$.
The important thing here to note is that $c$ is independent of $F_X$ so $c$ does not depend on the unknown $F_X$.
In your specific case $C_n(F)$ does not depend on $F$ (as long as $F$ is continuous I would guess), so the distribution function $G$ of $C_n(F)$ is also independent of $F$.
Moreover in this specific case $G$ is also continuous.
Therefore you can pick $c$ (irrespective of $F$) as a quantile of $G$ to get
$$
\mathbb{P}(C_n(F) > c) = \alpha.
$$
