In Georgii's book they state:
Given $(\chi,\mathcal{F},P_0,P_1)$ with simple hypothesis and alternative and $0<\alpha<1\ $ a given significance level. Then:
$(a)$ There exists a Neyman-Pearson-test $\phi$ with $E_0(\phi)=\alpha$.
They begin the proof like this:
Let $c$ be any $\alpha$-fractil ($1-\alpha$ quantil) of $P_0\circ R^{-1}$ where $R$ is the likelihood-quotient. By definition we have: $$P_0(R>c)\leq \alpha \quad P_0(R\geq c)\geq \alpha$$ Exactly this is the part I don't get. How come we have $P_0(R\geq c)\geq \alpha$? $P_0(R>c)\leq \alpha$ is exactly by definition of the $1-\alpha$-quantil but how come taking one more point into consideration we have that it becomes greater than $\alpha$. Could anyone explain to me how that is derived?
Here are some details. Define $\beta(x)=P_0(R>x)$, so that $1-\beta(x)$ is a c.d.f., thus non-decreasing and right-continuous. Thus, given $\alpha\in(0,1)$, there is $c$ s.t. $$1-\beta(c^-)\le 1-\alpha\le 1-\beta(c),$$ where $c^-$ denotes the left limit. Using these inequalities, $P_0(R>c)=\beta(c)\le\alpha$ and $$P_0(R\ge c)=\beta(c^-)\ge \alpha.$$