The scalar quantum field operator is defined
$$\phi_0(\vec x,t) = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}} \left(a_p e^{-ipx}+a_p^{\dagger}e^{ipx}\right)$$
in Schwartz eq. 2.78. Here $\omega_p = \sqrt{\vec{p}^2+m^2}$. One can show that
$$\int \frac{d^3\vec{k}}{2\omega_k} = \int d^4k \delta(k^2-m^2)\theta(k^0)$$
is a Lorentz invariant measure. I don't see how the measure in the definition of the quantum field is Lorentz invariant because of that square root.
We want to construct the field such that $$ \langle\vec{p}|\phi(x)|0\rangle=e^{-i\vec{p}\cdot \vec{x}} $$ and the definition is $\langle \vec{p}|=\langle 0|\sqrt{2\omega_p} a_p$, i.e. the state $|\vec{p}\rangle$ have an invariant inner product: $$ \langle \vec{k}|\vec{p}\rangle=2\omega_p (2\pi)^3\delta^3(\vec{p}-\vec{k}) $$ such that the identity operator can be writen as: $$ \mathcal{1}=\int \frac{d^3 p}{(2\pi)^3}\frac{1}{2\omega_p}|p\rangle\langle p| $$ so the square root is just cancelling the square root of the $\langle \vec{p}|$ state.
All this is convention. The Weinberg texbook chooses a different one in witch $$ \langle \vec{k}|\vec{p}\rangle= (2\pi)^3\delta^3(\vec{p}-\vec{k}) $$ and the price is to introduce a new factor into the Lorentz transformation of the state $|\vec{p}\rangle$: $$ U(\Lambda)|\vec{p}\rangle = \sqrt\frac{(\Lambda p)^0}{p^0} |\vec{p}_{\Lambda}\rangle $$
The measure you consider is not Lorentz invariant. The point is that, under Lorentz transformations, $a_p$ and $a_p^\dagger$ are not $p$-scalar fields but they take a factor which compensates the failure of the measure to be Lorentz invariant, and the integral defining the quantum field produces a scalar field. Nogueira's answer includes all information necessary to write down the transformation rule of $a_p$ and its Hermitian conjugate companion. However all that is matter of convention, since one could use the invariant measure from scratch.
