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I want to show that if $f:X\rightarrow\mathbb R$ is measurable and $a>1$, then $f$ is integrable iff $$\sum_{n\in\mathbb {Z}}a^n\mu(\{a^n\le\lvert f\rvert < a^{n+1}\})<\infty.$$ If we define $A_n=\{a^n\le\lvert f\rvert < a^{n+1}\}$, which are disjoint, measurable sets then of course $s_n:=\sum_{-n}^n a^k\chi_{A_k}$ are nice simple functions for all $n\in\mathbb Z$. However $f$ is not dominated by $s_n$ and $s_n$ don't converge to $f$, so I can't use the go-to convergence theorems.

This also reminds me of Borel-Cantelli from probability theory, but this is a problem from measure/integration theory.

Is there some way to derive the integrability of $f$ from the integrability of the simple functions? Or am I missing a better approach?

Ruben Kruepper
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  • Almost the same: https://math.stackexchange.com/questions/182527/equivalence-of-a-lebesgue-integrable-function – Tito Eliatron Feb 19 '21 at 09:22
  • You want $a>1$, not just $a>0$, otherwise that sum over $\Bbb Z$ is zero. This reminds me more of Cauchy condensation, the only difference being that $a^n\mu{a^n\le \lvert f\lvert< a^{n+1}}$ isn't decreasing. –  Feb 19 '21 at 09:25
  • @TitoEliatron I see what you mean, but that "theorem" was in fact wrong (going off of the accepted answer). – Ruben Kruepper Feb 19 '21 at 09:25
  • @Gae.S. yes you're right, I mistyped – Ruben Kruepper Feb 19 '21 at 09:25
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    Also, $\sum_{k\in\Bbb Z}a^k\chi_{A_k}$ isn't simple, because its range is countable, but not necessarily finite. –  Feb 19 '21 at 09:27

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We have $\int |f|d\mu=\sum \int_{A_n} |f|d\mu$ and

$$a^{n}\mu (A_n) \leq \int_{A_n} |f|d\mu \leq a^{n+1}\mu (A_n)$$

That is all you need. No theorems like Borel-Cantelli , DCT etc are required.

Kavi Rama Murthy
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  • I see you're argument, but my issue is that we make a statement about the sum of the left inequality. So $\sum a^n\mu(A_n)<\infty$ does not necessarily mean $\sum\int_{A_n}\lvert f\rvert d\mu < \infty$ – Ruben Kruepper Feb 19 '21 at 09:29
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    @RubenKruepper $\sum a^{n}\mu(A_n) <\infty$ implies that $\sum a^{n+1}\mu(A_n) =a\sum a^{n}\mu(A_n) <\infty$ so the right hand inequality implies that $\int |f|d\mu <\infty$. – Kavi Rama Murthy Feb 19 '21 at 09:31