From $\left\lVert \sup_{M>M_0} \left| \sum_{k=M_0}^M f_k \right| \right\lVert_2 < \epsilon$ show convergence a.e. of the series. I'm having trouble with the following 'qual' problem. For one, I don't know what to make of the absolute value inside the $L^2$-norm. In short, I just don't have any intuition for it. And I don't understand some steps in the official solution. Could you please help me get a feel for it or understand those steps? The statement is this:

Let $(X,\mathcal{A},\mu)$ a finite measure space and $\{f_k\}_1^{\infty}$ a sequence of square-integrable functions with the following property: for all $\epsilon > 0$ there exists and $M_0\in \mathbb{N}$ so that $$\left\lVert \sup_{M>M_0} \left| \sum_{k=M_0}^M f_k \right| \right\lVert_2 < \epsilon.$$ Show that the series $$\sum_1^{\infty}f_k$$ converges a.e.

The official solution is here: http://s22.postimg.org/fmdq04nox/x_29xdis82.jpg
I don't understand that bit about choosing numbers $n_1$, $n_2$... And I'd like to know how to think about this problem in a way that would allow me to solve it by myself.
Thanks!
 A: To prove the a.e. convergence, the method we think about may be the squeeze theorem. That is, we find two functions $F^*$ and $F_*$ that bounds the limit function. If the upper and lower bound are the same a.e., we can say that the limit exists a.e.. The natural way to choose the upper and lower bound, as we can see, is choosing 
$$F^*=\limsup F_n,F_* = \liminf F_n$$
where $F_n$ is the partial sum.
The next thing we would like to prove is $a.e$ equality between upper and lower bound. The way is to use the definition, say, the points where the upper and lower bound are not equal is of measure zero. The set to define the points with "not equal upper-lower bound" are the union of those points that upper bound and lower bounds differs at least $2^{-t}$, where $t$ varys from $1$ to infinity that take all positive integer value. This set 
$$\bigcup_{t=1}^\infty \{x|\forall M>0,\exists n_1,n_2>M,\text{ such that } F_{n_1},F_{n_2}\text{ satisfies }|F_{n_1}-F_{n_2}|>2^{-t}\}=\bigcup_{t=1}^\infty E_t.$$
is exactly the set we want. So we observe each set, when we prove that each of them are of measure zero, by subadditive we know that the total measure is zero. 
Finally, we make use of our condition to prove that each of these sets are measure zero. This follows from our definition of these set when using the integration. Since for each given $M_0$ there's two functions indexed by $n_1,n_2>M_1$ that differs at least $2^{-t}$. 
I think the official answer here makes a mistake. Suppose we take $F_k(x)=(-1)^k 2^{-t-1}\cdot 1.1$ when $k>M_0$ and $F_{M_0}(x)=0$, then $x\in E_t$ but $\sup_{M>M_0}|F_{M}-F_{M_0}|<2^{-t}$ holds anyway. But the problem can be resolved using a minor change. $x\in E_t$ implies $\sup_{M>M_0}|F_{M}-F_{M_0}|>2^{-t-1}$. Since if both $F_{n_1}$ and $F_{n_2}$ differs from $F_{M_0}$ less than $2^{-t-1}$, the result is they differs each other less than $2^{-t}$, then $x\not\in E_t$. Thus the inclusion $E_t \subset F_t=\{x|\sup_{M>M_0}|F_{M}(x)-F_{M_0}(x)|>2^{-t-1}\}$ implies
$$\mu(E_t)=\int_{E_t}1 d\mu \le\int_{F_t}1 d\mu .$$
Chebyshev inequality gives us 
$$\int_{F_t}1 d\mu\le 2^{2t+2}\int_X \left(\sup_{M>M_0}|F_{M}(x)-F_{M_0}(x)|\right)^2 d\mu\le 2^{2t+2}\epsilon^2.$$
Use arbitrainess of $\epsilon$, we know that the set we consider at previous step has measure $0$, which finishes the proof.
