Question above proof of Marcinkiewicz Interpolation Theorem

I am reading through the proof in Tao's lecture notes and I have a question about justifying a "without loss of generality" statement.

Statement: Let $T$ be a sublinear operator taking functions on $(X,\mathcal{B}_X,\mu)$ to functions on $(Y,\mathcal{B}_Y,\nu)$. Let $0 < p_o, p_1, q_0, q_1 \leq \infty$ and $A_0, A_1 > 0$ be such that $\|Tf\|_{q_i,\infty} \lesssim_{p_i,q_i} A_iHW^{\frac{1}{p_i}}$ for all sub-step functions of height $H$ and width $W$ and $i = 1,2$. Then for any $\theta \in (0,1)$ and $r \in [1,\infty]$ with $q_\theta > 1$, we have $$\|Tf\|_{q_\theta,r} \lesssim_{p_0,p_1,q_0,q_1,r,\theta} A_{\theta} \|f\|_{p_\theta,r}$$ for all simple function $f$ with finite measure support where $\frac{1}{p_\theta} := \frac{1-\theta}{p_0} + \frac{\theta}{p_1}, \frac{1}{q_\theta} := \frac{1-\theta}{q_0} + \frac{\theta}{q_1},$ and $A_\theta := A_0^{1-\theta}A_1^\theta$.

My question is about the following reduction. We observe the following symmetries. First, we can multiply $T$ (and the $A_\theta$) by an arbitrary constant. Moreover, we may multiply the underlying measure $\mu$ by a constant $C$ and $A_\theta$ by $C^{\frac{-1}{p_\theta}}$. Similarly, we may multiply the underlying measure $\nu$ by a constant $C$ and $A_\theta$ by $C^{\frac{1}{q_\theta}}$. Therefore, we may assume $A_0 = A_1 = 1$ (and therefore $A_\theta = 1$).

Usually, when I am trying to justify a "without loss of generality" claim, I assume the more specific claim is true, i.e., the claim is true for $A_0 = A_1 = 1$, and try to generalize. So assume it's true in the special case. I can apply the special case of the theorem to the operator $aT$ using the measures $b\mu$ and $c\nu$ for some parameters $a,b,c$. Then I have $\|aTf\|_{q_i,\infty} \lesssim_{p_i,q_i} ab^\frac{1}{p_i}c^\frac{-1}{q_i}A_iHW^{\frac{1}{p_i}}$ for all sub-step functions of height $H$ and width $W$ and $i=0,1$. Now in order to apply the special case of theorem, I need to select $a,b,c$ such that $ab^\frac{1}{p_i}c^\frac{-1}{q_i}A_i = 1$. It seems to me that scaling $T$ by $a$ is unnecessary. If we just select $a=1$ and $b,c$ such that $b^\frac{1}{p_i}c^\frac{-1}{q_i}A_i = 1 \,\,\, (i=1,2)$, then we can use the special case of the theorem to conclude $$\|Tf\|_{q_\theta,r} \lesssim_{p_0,p_1,q_0,q_1,r,\theta} \|f\|_{p_\theta,r}$$ Now if we rescale to the original measures, $b\mu \mapsto \mu$ and $c\nu \mapsto \nu$ we get $$\|Tf\|_{q_\theta,r} \lesssim_{p_0,p_1,q_0,q_1,r,\theta} b^\frac{-1}{p_\theta}c^\frac{1}{q_\theta}\|f\|_{p_\theta,r} = A_{\theta}\|f\|_{p_\theta,r}$$

Is this reasoning correct? Is the homogeneity with repsect to scaling $T$ unnecessary?