Convergence a.e. of series $\sum_{n=1}^\infty c_n\phi_n$ where $(\phi_n)_n$ is orthonormal and $(c_n)_n \in \ell^p$ I'm trying to understand a paragraph from the article Christ, Kiselev: Maximal functions associated to filtrations. After the proof of Theorem 1.1., the authors deduce a theorem of Menshov as a corollary:

Let $(X, \mu)$ be a measure space and let $(\phi_n)_n$ be an orthonormal sequence in $L^2(X)$. Let $1 \le p < 2$. Then for every sequence $(c_n)_n \in \ell^p$ the series
$$\sum_{n=1}^\infty c_n\phi_n$$
converges a.e. in $X$.

Obviously the series $\sum_{n=1}^\infty c_n\phi_n$ converges in $L^2(X)$ due to Parseval since in particular we have $(c_n)_n \in \ell^2$.
Theorem 1.1. from the article gives that the (sublinear) map $$\ell^p \to L^2(X), \quad c = (c_n)_n \mapsto \sup_{N\in\Bbb{N}} \left|\sum_{n=1}^N c_n\phi_n\right|$$
is well-defined and bounded in the sense that there exists a constant $C>0$ such that
$$ \left\|\sup_{N\in\Bbb{N}} \left|\sum_{n=1}^N c_n\phi_n\right|\right\|_{L^2(X)} \le C\|(c_n)_n\|_p, \quad \text{ for all }(c_n)_n \in \ell^p.$$
The authors now conclude that the statement of the theorem clearly follows from this fact and from the fact that $\sum_{n=1}^\infty c_n\phi_n$ obviously converges a.e. when $(c_n)_n$ is finitely-supported.
I'm not sure how it follows. I noticed that for $M \ge N$ the Cauchy sums are also bounded by the supremum by reverse triangle inequality:
$$\left|\sum_{n=N+1}^M c_n\phi_n\right| \le 2\sup_{K\in\Bbb{N}} \left|\sum_{n=1}^K c_n\phi_n\right|.$$
So this is bounded a.e., however we would need that it goes to $0$ as $M,N\to\infty$. Am I missing something easy?
 A: This is a standard argument in harmonic analysis. Let me give the abstract version first.
Suppose we have a sequence of linear operators $T_n$ mapping measurable functions on $(X, \mu)$ to measurable functions on $(Y,\nu)$ such that
$$
T_*h(x) := \sup_{n} |T_nh(x)| 
$$
satisfies $T_* :L^p(X)\to L^q(Y)$, for some $1\leq p,q<\infty$. Suppose also that for $h$ in a dense set $E\subset L^p(X)$ it holds that $(T_nh(y))_n$ is a Cauchy sequence for a.e. $y \in Y$. Then $(T_nh(y))_n$ is a Cauchy sequence for a.e. $y$ and all $h\in L^p(X)$.
In other words, pointwise control on a dense subset, plus some control of the maximal operator $T_*$ imply pointwise control for all admissible functions. In fact you can get away with weaker control on $T_*$ than the above $L^p$ bounds, but I won't get into that.
The proof is very simple: Given $h\in L^p(X)$ and $\varepsilon>0$ fixed, let $g\in E$ be such that $\| h-g\|_{L^p(\mu)}^{p/q}$ small, then by properties of $\limsup$ we have
\begin{equation}
\begin{split}
\nu(\{ y\in y: \limsup_{n,m} |T_n h(y)-T_mh(y)|>\varepsilon\}) & \leq \mu(\{ \limsup_{n}|T_nh(y)-T_n g(y)|>\varepsilon/3\}) \\
& \quad + \mu(\{ \limsup_{n,m}|T_n g(y)-T_mg(y)|>\varepsilon/3\}) \\
& \quad + \mu(\{ \limsup_{m}|T_m g(y)-T_mh(y)|>\varepsilon/3\})\\
& =: I+II+III.
\end{split}
\end{equation}
Now, $I$ and $III$ we can estimate by Chebyshev's inequality as
$$
\mu(\{ \limsup|T_n h(y)-T_n g(y)|>\varepsilon/3\}) \leq \dfrac{3}{\varepsilon} \int_Y T_*(h-g)(y)^q\, d\nu(y) \lesssim \dfrac{1}{\varepsilon} \| h-g\|_{L^p(\mu)}^{p/q},
$$
by the boundedness of $T_*$. If we choose then $f-g$ close enough, then $I$ and $III$ can be made as small as we want, while $II=0$ always by hypotheses (since $g\in E$). This means that the set where $ \limsup_{n,m}|T_nh(y)-T_mh(y)|>\varepsilon/3$ has measure 0 for all $\varepsilon>0$. This implies that $T_n h$ is Cauchy.
In your case you operators are mapping $\ell^p$ to $L^2(X)$ via
$$
T_N: (c_k)_k \mapsto \sum_{k=1}^N c_k\phi_k,
$$
so that $T_*$ is exactly the operator you have bounds on, and the result follows from the above abstract principle.
A: For each $m \in \mathbb{N}$, by applying the inequality to the residual tail $n \mapsto c_n \mathbf{1}_{\{n\geq m\}},$ we get
$$ \left\| \sup_{N\in\mathbb{N}} \left| \sum_{n=m}^{N} c_n\phi_n \right| \right\|_{L^2(X)} \leq C \left\| (c_n \mathbf{1}_{\{n\geq m\}})_{n\in\mathbb{N}} \right\|_{p}. $$
In particular,
$$ \sup_{N\in\mathbb{N}} \left| \sum_{n=m}^{N} c_n\phi_n \right| \quad \xrightarrow[m \to \infty]{\text{in } L^2(X)} \quad 0, $$
and then we can extract a subsequence $(m_k)_{k\in\mathbb{N}}$ such that
$$ \sup_{N\in\mathbb{N}} \left| \sum_{n=m_k}^{N} c_n\phi_n \right| \quad \xrightarrow[k \to \infty]{\text{a.e. in } X} \quad 0. $$
This then guarantees the a.e. convergence of the sum $\sum_{n=1}^{\infty} c_n \phi_n$.
