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Let $(X,\mathcal{A},\mu)$ be a measure space and $f: X\rightarrow \mathbb{R}$ measurable. $f$ is integrable if and only if $\sum\limits_{n \in \mathbb{Z}}a^n\mu (\{a^n\leq|f|.

My attempt was to use the Tchebycheff Inequality: $\mu (\{f\geq a\} \cap E)\leq \frac{1}{a}\int f d\mu$, but it did not work.

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    You mean "$f$ is integrable if ..." instead of measurable, I suppose – 2017-01-26
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    Yes of course, i have changed it – 2017-01-26

2 Answers 2

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As $f$ is measurable and $a>1$, the function $$g(x)=\begin{cases}0&f(x)=0\\a^{\lfloor \log_a|f(x)|\rfloor }&f(x)\ne 0\end{cases}$$ is also measurable: For $c>0$, we have $\mu(g\ge c)=\mu(a^{\lceil \log_ac\rceil }\le |f|)$. And of course $g$ is non-negative.

Note that $$ g(x)\le |f(x)|\le ag(x).$$ Hence if $g$ is integrable then also $\int f\,\mathrm d\mu\le a\int g\,\mathrm d\mu<\infty$ and so $f$ is integrable. If on the other hand $f$ is integrable, then $\int |f|\,\mathrm d\mu <\infty$ and also $\int g\,\mathrm d\mu\le \int |f|\,\mathrm d\mu<\infty$.

Thus $f$ is integrable iff $g$ is. But $$\int g\,\mathrm d\mu = \sum_{n\in\Bbb Z}a^n\mu(\{\,a^n\le |f|

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    I do not understand why $\int g d\mu = \sum_{n \in \mathbb{Z}} a^n\mu ({a^n\leq |ƒ| < a^{n+1}})$ – 2017-01-26
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    @user408161 : Note that $g$ is constant in a set of the form $\{a^n\le|f|< a^{n+1}\}$, and its value is $a^n$. – 2017-01-26
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Note that $$a^n \chi_{\{a^n \le \lvert f \rvert < a^{n+1}\}} \le \lvert f \rvert\chi_{\{a^n \le \lvert f \rvert < a^{n+1}\}} < a^{n+1}\chi_{\{a^n \le \lvert f \rvert < a^{n+1}\}}.$$ Since the sets from a disjoint cover of $X$, summing the above gives $$\sum_{n\in \mathbb Z}a^n \chi_{\{a^n \le \lvert f \rvert < a^{n+1}\}} \le \lvert f \rvert < \sum_{n\in \mathbb Z}a^{n+1}\chi_{\{a^n \le \lvert f \rvert < a^{n+1}\}}.$$ Now just integrate both sides over $X$ and use the monotone convergence theorem to integrate term-by-term in the sums.