# Convergence in mean with additional hypothesis implies convergence in measure

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.

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.

• $\sum_{n}\|f_n - f\|_{L^1} = \sum_{n}\int|f_n - f| = \int \sum_{n}|f_n - f| < \infty \implies \sum_{n}|f_n - f| < \infty$ a.e. Nov 9, 2022 at 7:01

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.