What is the common $t$-test for $H_0: \mu = \mu_0 \ \text{ and } \ H_A:\mu > \mu_0$? The following remark came up in my lecture:

Let $X_1,X_2,\ldots,X_n$ be normally distributed with known variance $\sigma^2 > 0$.
For testing
$$H_0: \mu = \mu_0 \ \text{ and } \ H_A:\mu > \mu_0.$$
we use the "common level-$\alpha$-test". (If I am not mistaken my teachter meant the $t$-test with this.)

In the lecture we only did tests where the alternative hypothesis was the complement of the hypothesis, so I do not understand what the "common level-$\alpha$-test" is supposed to be. Could you please explain this to me?
To clearify: My problem here is that I do not understand how to formulate a test where $H_A$ is not complementary to $H_0$.
 A: You are correct; technically speaking, the null hypothesis and alternative hypothesis are complementary, i.e. they specify regions that are mutually exclusive and collectively exhaust the parameter space (i.e. they partition the space).
The reason you may see the composite null hypothesis in a one-sided test $H_0: \mu\leq \mu_0$ replaced with a simple null hypothesis $H_0: \mu=\mu_0$ has to do with how we conduct tests. The aim is to control the size of the test, which is the supremum of the type 1 error over all data generating processes satisfying the null hypothesis:
$$\sup_{h \in H_0}P(\text{reject } H_0|h)\leq \alpha.$$
That is, we want to ensure the "worst case" type 1 error is below cutoff $\alpha$. For the composite null $H_0: \mu\leq \mu_0$, the worst case type 1 error corresponds with $\mu=\mu_0.$ Visually you can see this by drawing any other normal curve with mean $\mu<\mu_0$ and noting that the probability of rejecting the null (right tail area above some cutoff) is lower than in the case $\mu=\mu_0$.
In summary, it is without loss to replace $H_0:\mu\leq \mu_0$ with $H_0:\mu=\mu_0$ in a one-sided t test; just treat them the same and carry on.
