Let $ a = \{a_n \}_{n=1}^{+\infty}$ be a decreasing summable sequence with $a_i >0$ for all $i$, that is, $\sum_{n \geq 1} a_n < \infty$ with $a_i \geq a_{i+1} $ for all $i$.
The definition of the cut-out set $C_a \subset \mathbb{R}$ defined by $a$ is as follows. (This definition can be found in arXiv:1604.01234v1)
In the first step, we remove from $I = [0, \sum_{n=1}^{\infty} a_n]$ an open interval of length $a_1$, resulting in two closed intervals $I_1^1$ and $I_2^1$. Having constructed the $k$-th step, we obtain the closed intervals $I_1^k, \dots , I_{2^k}^k$ contained in $I$. The next step consists in removing from each $I_j^k$ an open interval of length $a_{2^k+j-1}$, obtaining the closed intervals $I^{k+1}_{2j−1}$ and $I^{k+1}_{2j}$. Then $$C_a = \bigcap_{k \geq 1} \bigcup_{j=1}^{2^k} I^k_j$$ is called the the cut-out set defined by $a$.
For the given sequence $a$, I see that $C_a$ is unique.
Next, given $a$, We call $$s_n = \frac{1}{2^n}\sum_{j=2^n}^{\infty} a_j$$ the average of $n$ and $I_j^{n}$ be the $j$-th interval arising in the $n$-th step of the cut-out set $C_a$.
My question is, why this inequality $$s_{n+1} \leq |I_j^n| \leq s_{n-1} \quad \forall 1 \leq j \leq 2^n$$ holds for all $n$ under the assumption that $a$ is a decreasing summable sequence?