In this post, I have provided a general framework of hypothesis testing.
The classical loss function associated with hypothesis testing is
\begin{align}L(\theta, a_0) &= \mathbb I_{\Theta_1}(\theta)=\begin{cases}1, &\theta\in\Theta_1,\\0,&\theta\in\Theta_0,\end{cases}\\ L(\theta, a_1) &= \mathbb I_{\Theta_0}(\theta)\end{align}
and the corresponding risk follows as
$$R(\theta,\delta) = \begin{cases}\alpha_\delta(\theta) := \mathbb E_\theta\varphi(\mathbf X), &\theta\in\Theta_0\\\beta_\delta(\theta) := 1-\mathbb E_\theta\varphi(\mathbf X), &\theta\in\Theta_1\end{cases}.\tag{1}$$
What do we seek to do in hypothesis testing? Well precisely, we need to find $\delta$ in such a way as to minimize $R(\theta_1,\delta)=1-\mathbb E_{\theta_1}\varphi(\mathbf X),$ for all $\theta_1\in\Theta_1$ subject to the constraint $R(\theta_0,\delta)=\mathbb E_{\theta_0}\varphi(\mathbf X)\leq\alpha$ for all
$\theta_0\in\Theta_0.$ What does the constraint mean? Again writing it:
$$\forall\theta\in\Theta_0,~~\mathbb E_\theta\varphi(\mathbf X)\leq \alpha\implies \sup_{\theta\in\Theta_0} \mathbb E_\theta\varphi(\mathbf X)=\alpha.$$ Why the equality in place of $\leq$? Notice $\alpha$ is an upper bound, so the supremum has to be lesser than or equal to $\alpha.$ But the size is defined to be the equality case (as utobi mentioned in the comment, level is defined with $\leq$ constraint).
As for your argument on reasoning that the expectation of a decision rule is a statistic and all that, I couldn't make any sense. But the motivation for why supremum is introduced is given above.