See for example A. E. Ingham, The distribution of prime numbers (1932), Chapter IV: Explicit formulae.

Here's another way to get the lower bound of $\lVert A \rVert^n \gg \lvert\det(A)\rvert \gg p^{n-k+1}$. Namely, take a nonsingular $A$ of rank $\leq k-1$ and let $L$ be the lattice $A \mathbb Z^{n}$. Then $L \subset p\mathbb Z^{n} + \text{span of }k-1 \text{ vectors}$, so that $[0,p]^n$ is cut by $L$ into pieces of volume roughly $\text{Vol}([0,p])^n / p^{k-1}$ and you get the lower bound on $\text{Vol}(L)$ you want.

Here is a paper that uses the terminology in Matt Young's comment: arxiv.org/abs/math/0206018 (search for "critical value").

Does Lemma 2.2 in [3] or its proof not answer your question?

@BCLC Yes, different person.

This is very similar to "convergence in $L^1$ $\implies$ convergence almost everywhere?", and it seems you can use this to produce counterexamples. FWIW, one situation where I think the implication holds, is when $\Omega$ has a nice topology, $E_1 = \mathbb C$, $f$ is jointly continuous and $L^p$-holomorphic. When you then write $f(u, \omega) - f(0, \omega) = u f'(0, \omega) + u R(u, \omega)$; you can use Cauchy's integral formula to prove that $f'$ and $R$ are jointly continuous, and then $R \to 0$ in $L^p$ implies $R \to 0$ everywhere; i.e. $f$ is pointwise holomorphic (=for fixed $\omega$).

@FrançoisG.Dorais $\mu(z\Sigma) = r^2 \mu(\Sigma)$, right? So that gives $r \geq 1/\sqrt 2$.