Question about a square integrable uniformly bounded orthonormal sequence $(e_n)$ such that $\sum c_n e_n$ converges a.e.

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This is part of the proof of theorem 28.27 from Rene Schilling's Measures, Integrals and Martingales.

Let $(X, \mathscr{A},P)$ be a probability space and $(e_n)_{n \in \mathbb{N}_0} \subset L^2(P)$ be independent random variables such that $E(e_n)=0$ and $E(e_n^2) = 1$ and let $(c_n)_{n \in \mathbb{N}_0}$ be a sequence of real numbers.

Suppose also that $\sup_{n \in \mathbb{N}_0}\Vert e_n \Vert_\infty \le \kappa < \infty$ and $\sum_{n=0}^\infty c_n e_n$ converges almost everywhere.

Now let $u_n:= \sum_{k=0}^n c_k e_k.$

Consider the stopping time $\tau = \tau_\gamma := \inf \{n \in \mathbb{N}_0: |u_n| > \gamma\}, \inf \emptyset = \infty.$ In this situation, how can we choose $\gamma >0 $ in such a way that $$\kappa^2 P(\tau < \infty) < \frac{1}{2} P(\tau= \infty)$$

since the series $\sum_{i=0}^\infty c_i e_i$ converges a.e.? And how do we get $P(\tau = \infty) > 0 $ for sufficiently large $\gamma$? For each $x \in X$ for which $\sum c_n e_n(x)$ converges, we can find a $\gamma$ that bounds all $|u_n|$, so $\tau_\gamma = \infty$. However, I cannot see how to come up with a bound $\gamma$ that works for all $x$ in some subset of positive measure. I would greatly appreciate some help in establishing the above inequality.

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As you say, for almost every $x \in X$, there is a $\gamma$ such that $\tau_{\gamma}(x) = + \infty$. This means $$ \mathbb{P} \left ( \bigcup_{\gamma \in \mathbb{N}} \{ \tau_{\gamma} = + \infty \} \right ) = 1. $$ But as this is an increasing sequence, then $$ \lim_{n \to + \infty} \mathbb{P} \left ( \{ \tau_{\gamma} = + \infty \} \right ) = 1. $$ You can therefore find $\gamma$ such that $$ \mathbb{P} \left ( \{ \tau_{\gamma} = + \infty \} \right ) > \frac{\kappa^2}{\kappa^2 + 1/2}, $$ and rearranging gives you the equality you desire.

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The events $A_\gamma = \{\tau_\gamma = \infty\}$ are increasing with respect to $\gamma$ and, by your own element-wise analysis, $P\left(\bigcup_{\gamma > 0} A_\gamma\right) = 1$.

By continuity, $$\lim_{\gamma \to\infty} P(\tau_\gamma = \infty) = \lim_{\gamma \to \infty} P(A_\gamma) = 1$$