Lebesgue measure and Rational enumeration in $[0,1]$

Question

Let $q_1,q_2,\dots$ be an enumeration of all the rationals in $[0,1]$. Define function $f(\omega) = \sum\limits_{n=1}^{\infty} 2^{-n}|\omega-q_n|^{-1/3}$. Prove that $\int_{[0,1]}f(\omega)m(d\omega)<\infty$ where $m$ is the Lebesgue measure.

I try to split $[0,1]$ to rational part (measure 0) and irrational part(measure 1). For irrational part, I hope function $f(\omega)$ is bounded then I am done. But I don't see how to bound it, since the power is $-1/3$.

score 1 · Accepted Answer · answered Sep 23 '20 at 05:16

1

To avoid doing much calculus, we estimate $$\int_{[0,1]}|\omega-q_n|^{-1/3}=\int_{[-q_n,1-q_n]}|\omega|^{-1/3}\leq\int_{[-1,1]}|\omega|^{-1/3}=2\int_0^1\omega^{-1/3}d\omega=3.$$ (Of course you could have computed this exactly but that would be annoying.) Therefore $$\int_{[0,1]}f\leq\sum_{n=1}^\infty2^{-n}\cdot3=3.$$

answered Sep 23 '20 at 05:16

Funktorality

3,221

What happens if $\omega=q_k$ for some $k$? I don't think we can compute the integral. – user387147 Sep 23 '20 at 18:48
@user387147 the above computation still holds because $|\omega|^{-1/3}$ is integrable on $[-1,1]$. – Funktorality Sep 23 '20 at 21:05

Lebesgue measure and Rational enumeration in $[0,1]$

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