Let $(X,\mathcal{A},\mu)$ be a measure space and $f: X\rightarrow \mathbb{R}$ measurable. $f$ is integrable if and only if $\sum\limits_{n \in \mathbb{Z}}a^n\mu (\{a^n\leq|f|<a^{n+1} \})<\infty$.

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.

  • $\begingroup$ You mean "$f$ is integrable if ..." instead of measurable, I suppose $\endgroup$ – Hagen von Eitzen Jan 26 '17 at 15:08
  • $\begingroup$ Yes of course, i have changed it $\endgroup$ – user408161 Jan 26 '17 at 15:19

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 $$\int g\,\mathrm d\mu = \sum_{n\in\Bbb Z}a^n\mu(\{\,a^n\le |f|<a^{n+1}\,\}).$$

  • $\begingroup$ I do not understand why $\int g d\mu = \sum_{n \in \mathbb{Z}} a^n\mu ({a^n\leq |ƒ| < a^{n+1}})$ $\endgroup$ – user408161 Jan 26 '17 at 15:47
  • 1
    $\begingroup$ @user408161 : Note that $g$ is constant in a set of the form $\{a^n\le|f|< a^{n+1}\}$, and its value is $a^n$. $\endgroup$ – dafinguzman Jan 26 '17 at 16:43

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.