I would like to know how to derive the empirical expression of the test statistics below: $$ \omega^2 = \sum_{i=1}^{n}(U_{(i)}-\frac{2i-1}{2n})^2 + \frac{1}{12n} $$
from $$ \omega^2 = n \int_\Omega(F_n(x)-F(x))^2dF(x) $$
I presume that :
1.approximate $dF(x)=dF_n(x)=\frac{1}{n}$ and $dF_n(x)$ is discrete that $n \int_\Omega$ can be replaced to $\sum_{i=1}^{n}$
2.approximate $F_n(x)=\frac{2i-1}{2n}$. This is appropriate that $dF_n(x)=\frac{1}{n}$ and to prevent from $F_n(x_{(n)})=1$ that subtle adjustment declining $\frac{1}{2n}$ from each $F_n(x) $ might be added .
3.because $F(X) \sim U(0,1)$ , $F(x)$ can be expressed as $U$ and to correspond to each $F_n(x) $, $U$ should be orderd.
So, I could understand $n \int_\Omega(F_n(x)-F(x))^2dF(x)$ can be approximated as $\sum_{i=1}^{n}(U_{(i)}-\frac{2i-1}{2n})^2$, but I could not recognize how to derive the residual term $\frac{1}{12n}$.
