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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.

Julien
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Truong
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2 Answers2

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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).

Daniel Fischer
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  • Does convergence in $L^9$ imply pointwise convergence? – Truong Dec 21 '13 at 13:34
  • So maybe the answer is no – Truong 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? – Truong 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
  • I still dont get your idea. The sequence {$h_{n}$}$=\Sigma$${f_{i}}$ converge to $h$ implies that a subsequence {$h_{n_{k}} $} converge pointwise to $h$, but how can we conclude that the whole sequence is pointwise convergent? – Truong Dec 21 '13 at 13:48
  • Forget about $f_m$ for the moment. Consider only $g_m$ and $G_n$ first. From that you get the result about $\sum f_m$ with hint 3. – Daniel Fischer Dec 21 '13 at 13:50
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The answer is not in general. But you can always find an almost everywhere pointwise convergent subsequence.

See Does convergence in $L^{p}$ implies convergence almost everywhere?

Community
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Yiorgos S. Smyrlis
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  • 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