Proof of Neyman Pearson Lemma

Question

I am trying to understand the proof of Neyman Pearson Lemma as Uniformly Most Powerful test from here (Page 3).

It says the following:

Let $H_0: \theta = \theta_0$ and $H_a: \theta = \theta_1$. Also, let the test-statistic of NP be $\Lambda(x)$ and that of a different test be $K(x)$. For a given $\alpha$ level, let $c$ and $d$ be the critical values for $\Lambda(x)$ and $K(x)$ respectively.

Let \begin{align} & \text{A = [the likelihood-ratio test rejects $H_0$ and the other test does not],} \\ & \text{B = [the other test rejects $H_0$ and the likelihood-ratio test does not],} \\ \text{and } & \text{C = [both tests reject $H_0$]}\end{align}

It offers an simple proof as the following:

\begin{align} & \text{Pr}(A|H_1) = \int_A f(x | \theta = \theta_1) dx \gt \int_A c f(x | \theta = \theta_0) dx = c\text{Pr}(A|H_0) \\ \overset{?}{= } & \, \, c\text{Pr}(B|H_0) = \int_B c f(x | \theta = \theta_0) dx \gt \int_B f(x | \theta = \theta_1) dx = \text{Pr}(B|H_1)\end{align}

I am unable to see why the questionmarked equality is valid.

Answer 1

Both tests have the same significance level; assuming that they both attain that level exactly:

$P(\text{LRT rejects}|H_0)=P(\text{other test rejects}|H_0)=\alpha$

$\therefore P(\text{LRT rejects & other doesn't}|H_0)+P(\text{both reject}|H_0)\\\qquad=P(\text{other test rejects & LRT doesn't}|H_0)+P(\text{both reject}|H_0)$

$\text{or } P(\text{LRT rejects & other doesn't}|H_0)=P(\text{other test rejects & LRT doesn't}|H_0)$