Limit of the convolution of f and the approximation of unity

I have a question regarding the proof of this theorem (from the Textbook Real Analysis by E. M. Stein, R. Shakarchi)

Theorem 2.1If ${Kδ}_{δ}>0$ is an approximation to the identity and $f$ isintegrable on $\mathbb{R}^d$, then$$(f ∗ Kδ )(x) → f (x) \text{ as } δ→ 0$$for every $x$ in the Lebesgue set of $f$. In particular, the limit holds fora.e. $x$.

So the proof (I couldn't find it online), if anyone is familiar involves in writing the expression as

$$\int_{\mathbb{R^d}}{|f(x−y)−f(x)| |Kδ (y)| dy} =∫_{|y|≤δ}+∑^{\infty}_{k=0}∫_{2^k δ<|y|≤2^{k+1}δ}$$

(where I have omitted the RHS integral for formatting purposes)

So integrating on annuli of radius $2^k δ<|y|≤2^{k+1}δ$

and proceeding to use further properties of the Approximation to the Identity to estimate the second integral as $$c∑^{\infty}_{k=0}{\frac{1}{2^k}\frac{1}{{((2^k)δ)^d}}}\int_{|y|≤2^{k+1}δ}{|f(x−y)−f(x)| dy}$$where we use the fact that the measure of the ring is less than the measure of the whole ballSo my question is why are we specifically using Annuli with double the radius in our integrals and not just grabbing $δk $ and $δ(k+1)$, wouldn't then the estimation be the same?or is it just because we need the sum with $\frac{1}{2^k}$.

I hope I have been clear, I could understand if I wasn't.

I thank you all in advance.