Proving a seemingly simple property of Rademacher functions

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The Rademacher functions are given by $$r_k(t) = \mathrm{sgn}\sin(2^k\pi t), \quad t \in [0,1)$$ for each $k \in \mathbb{Z}_+$. These are basicaly "square-wave functions" of preiod $2,1,1/2, \dots$ and amplitude 1. These functions have many special properties, the main one being that they are independent in a probibalistic sense. I'm however interested in the following.

Let $a_1, a_2, \dots , a_N$ be a real finite sequence and $1 \leq n \leq N$, then $$|\sum_{k=1}^N a_kr_k(t)| \geq |a_n|$$ on a set of measure at least 1/2.

If one can show that $$\sum_{k \neq n} a_kr_k(t)$$ has the same sign as $a_nr_n(t)$ on such a set, then the result would follow by $$|\sum_{k=1}^N a_kr_k(t)| = |\sum_{k \neq n} a_kr_k(t)| + |a_nr_n(t)| \geq |a_n|.$$ If $n=1$ the result follows from the fact that $$\sum_{k=2}^N a_kr_k(t)$$ is non-negative on a set of measure 1/2 and negative on the other half and $r_1(t)=1$. Im not sure how to argue for the other cases however ($n>1$), and I do expect that there might be some simpler way of reasoning. Does anyone see how this could be fixed?

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You noted their probabilistic independence.

Given three independent distributions, each of which only takes a finite number of values,

$\mathbb{P}(A+B\geq0\;|\;C\geq0)=\mathbb{P}(A+B\geq0)$.

Exercise!

So for example,

let $\mathbb{P}(A_1+A_3\geq0)=p$

and suppose without loss of generality that $a_2\neq0$.

Then $\mathbb{P}(|A_1+A_2+A_3|\geq|A_2|)$

$\geq \mathbb{P}(A_1+A_3\geq0\;|\;A_2\geq0)\times\mathbb{P}(A_2\geq0)+\mathbb{P}(A_1+A_3<0\;|\;A_2<0)\times\mathbb{P}(A_2<0)$

$=p\times 1/2+(1-p)\times1/2$.