Convergence in mean with additional hypothesis implies convergence in measure

Question

This is a problem from Cohn's Measure Theory book:

Suppose that $(X, \mathscr A, \mu)$ is a measure space and that $f$ and $f_1, f_2, \ldots $ belong to $\mathscr L ^1 (X, \mathscr A, \mu , \mathbb R)$. Show that if $\{ f_n \}$ converges to $f$ in mean so fast that $\sum_n \lVert f_n -f \rVert _1 < +\infty$ then $\{ f_n \}$ converges to $f$ almost everywhere.

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Here's my attempt which might be helpful in proving something later on:

Let us take $\varepsilon >0$ and $m \in \mathbb N$. Then $\mu \left( \{ x \in X : |f_n (x) -f(x) | > \varepsilon \}\right)\le \frac{1}{\varepsilon} \lVert f_n - f \rVert _1$ for any $n \in \mathbb N$.

Thus, for any $N \in \mathbb N$, we have $\sum_{n=N}^{\infty} \mu \left( \{ x \in X : |f_n (x) -f(x) | > \varepsilon \}\right)\le \sum_{n=N}^{\infty} \frac{1}{\varepsilon} \lVert f_n - f \rVert _1$.

Since $\sum_n \lVert f_n -f \rVert _1 < +\infty$, the RHS of the aforementioned inequality can be made arbitrarily small, hence we can find $N_m \in \mathbb N$ such that for $N\ge N_m$, we have $\sum_{n=N}^{\infty} \mu \left( \{ x \in X : |f_n (x) -f(x) | > \varepsilon \}\right)\le \frac{\varepsilon}{2^m}$.

I need to find set $A$ of measure zero such that $\{ x \in X : \{ f_n (x) \} \text{ does not converge to } f(x)\}$ is contained in $A$. It feels like I need to make an $\varepsilon/2^m$ argument but I am unable to see how.

Any hints to complete the problem will be appreciated.

Answer 1

Let us set $g_n = \sum_{k}^n |f- f_k|$. By Fatou's Lemma, we get $$\int_X \sum_{k = 0}^\infty |f - f_k| d \mu = \int_X \liminf_n g_n d\mu \le \liminf_n \int g_n d \mu = \sum_{k = 0}^\infty \|f - f_k\|_1 < +\infty.$$ This implies that $$\sum_{k = 0}^\infty |f - f_k| \in \mathcal L^1$$ so that $$\left\{x \in X ~\bigg|~\sum_{k = 0}^\infty |f(x) - f_k(x)| = +\infty\right\}$$ has a measure zero. This implies that $f_k \to f$ $\mu$-a.e.