# Does $\sum_{m=1}^{\infty}f_{m}\left(x\right)$ converge for almost every $x$ in $X$?

Let {$f_{m}$} be a sequence of measureable real-valued functions in $\left(X,\mathrm{\mathcal{M}},\mu\right)$. Suppose ${\displaystyle \sum_{m=1}^{\infty}\left\{ {\displaystyle \int_{X}\left|f_{m}\right|^{9}d\mu}\right\} ^{\frac{1}{9}}}<\infty$. Does $\sum_{m=1}^{\infty}f_{m}\left(x\right)$ converge for almost every $x$ in $X$?

I have no idea. Help me some hints.

Thanks a lot.

Hint 1: The answer is yes. (I hope it isn't from a multiple choice test and I just gave the entire thing away.)

Hint 2: Consider the functions $g_m = \lvert f_m\rvert$ and $G_n = \sum\limits_{m=1}^n g_m$. Use Minkowski's inequality (the triangle inequality) and monotone convergence.

Hint 3: Absolute convergence implies convergence.

Let $S = \sum\limits_{m=1}^\infty \lVert f_m\rVert_9$. By assumption $S < \infty$, and for all $n$ we have $$\lVert G_n\rVert_9 \leqslant \sum_{m=1}^n\lVert g_m\rVert_9 = \sum_{m=1}^n \lVert f_m\rVert_9 \leqslant S.$$ $(G_n)$ is a monotone sequence of non-negative functions, and so is $(G_n^9)$, so the pointwise limit $G(x) = \lim\limits_{n\to\infty} G_n(x)$ exists ($[0,\infty]$-valued), and the monotone convergence theorem tells us $$\int_X G(x)^9\,d\mu = \lim_{n\to\infty} \int_X G_n(x)^9\,d\mu \leqslant S^9 < \infty.$$ Since $G^9$ is integrable, $N = \{ x : G(x) = +\infty\}$ is a $\mu$-null set. On $X\setminus N$ we have $\sum\limits_{m=1}^\infty \lvert f_m(x)\rvert < \infty$, that is, $\sum\limits_{m=1}^\infty f_m(x)$ converges absolutely almost everywhere. A fortiori, $\sum\limits_{m=1}^\infty f_m(x)$ converges almost everywhere (to a finite limit).

• Does convergence in $L^9$ imply pointwise convergence? – chuyenvien94 Dec 21 '13 at 13:34
• So maybe the answer is no – chuyenvien94 Dec 21 '13 at 13:34
• @chuyenvien94 Convergence in $L^9$ does not imply pointwise convergence, only that a subsequence exists that converges pointwise everywhere, but that is usually proved in the way I propose. The sequence to consider here has special properties that guarantee that in this case $L^9$ convergence implies pointwise convergence almost everywhere. – Daniel Fischer Dec 21 '13 at 13:37
• I need more details. Can you make your hints more precise? – chuyenvien94 Dec 21 '13 at 13:45
• I've added one more thing to the hints, I hope that helps. – Daniel Fischer Dec 21 '13 at 13:48

The answer is not in general. But you can always find an almost everywhere pointwise convergent subsequence.

• We are looking at the sum of a sequence where the sum of norms converges. That implies the pointwise a.e. convergence of the sum. – Daniel Fischer Dec 21 '13 at 14:02