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In a measure space $(X,\cal{A},\mu)$, suppose we are given functions $f$ and $f_1,f_2,\ldots$ all in $\cal{L}(X,\cal{A},\mu,\mathbb{R})$ such that a condition stronger than mean convergence follows, i.e., we assume that

$$\sum_{n=1}^\infty\int|f_n-f|d\mu<\infty$$

which as we know implies convergence of the general term for this series, which is to say that the $f_n$'s converge to $f$ in mean.

There is an exercise in "Measure Theory" book by Donald Cohn (Birkhäuser) which asks to prove that the above series converging implies almost sure convergence of the sequence of functions to the given $f$.

IMPORTANT: The book has made previously clear that "mere" convergence in mean without the aforementioned stronger condition DOES NOT suffice to prove almost sure convergence to $f$. All we can say is, e.g., that a subsequence of the $f$'s will do.

Can anyone be more specific as to how to prove this assertion? Normally I've been completing almost all Cohn's exercises but this one I'm stuck at (normally I'm more proficient, I guess it's the summer hot slowing down my --alleged-- reasoning capabilities).

Any advice or proof will be appreciated.

Warm regards to all math minds out there!

MSC

mosagepa
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1 Answers1

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Monotone convergence shows that $\int\sum|f_n-f|<\infty$. Hence $\sum|f_n-f|<\infty$ almost everywhere, hence $|f_n-f|\to0$ almost everywhere.

David C. Ullrich
  • 89,985
  • So basically, the trick was to interchange sum with integral, which is allowed since the positive term series has non decreasing partial sums, right? Thanks so much. – mosagepa Aug 07 '16 at 14:18