A randomized Neyman-Pearson test is of the form
$$ \phi(X_1,\ldots,X_n):= 1 \text{ if } \frac{p_1(X)}{p_0(X)} > c_0, \ q \text{ if } \frac{p_1(X)}{p_0(X)} = c_0, \ 0 \text{ if } \frac{p_1(X)}{p_0(X)} < c_0, $$
where $X:=(X_1,\ldots,X_n)$ and all $X_i$ are identically indipendent distributed. $p_0$ and $p_1$ are the probability distriubtions of the null and alternate hypothesis. $c_0$ and $q$ are defined such that \begin{align} E_{p_0} [\phi(X)] &= 1\cdot p_0(\phi(X)=1)+ q \cdot p_0(\phi(X)) \newline &= p_0(\frac{p_1(X)}{p_0(X)} > c_0) + q \cdot p_0(\frac{p_1(X)}{p_0(X)} = c_0) \newline &\stackrel{!}{=} \alpha \end{align}
, where $\alpha \in (0,1)$ and $E_{p_0}$ is the expected value of $X$ with respect to the $p_0$ distribution. The lemma of my professors lecture notes says the following: For ALL randomized tests $\tilde{\phi}$ with $E_{p_0}[\tilde{\phi}(X)]\stackrel{?}{\leq} \alpha$ it holds that $$E_{p_1} [\tilde{\phi}(X)] \leq E_{p_1} [\phi(X)].$$ I am familiar with all the terms of the NP-Test and I know how to apply it. I don't really understand what the interpretation of this theorem is. In particular ...
''ALL": Could somebody provide an example of a hypothesis where you show a concrete $\tilde{\phi}$ which is different of $\phi$ and then show that this lemma holds in provided example.
''$\stackrel{?}{\leq}$'': Why is $E_{p_0}[\tilde{\phi}(X)] \leq \alpha$ ? I though the way we construct a randomized hypothesis test $\tilde{\phi}$ is such that $E_{p_0}[\tilde{\phi}(X)]= \alpha$.