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Let $\{x_1, x_2, ...\}$ be all the rational numbers in $[0, 1]$. Take $\alpha \in ]-1, 0[$ and $0^\alpha = 1/0^{-\alpha} = 1/0 = \infty$ (I didn't write this, the teacher did, so please don't crucify me). We define $f$ from $[0, 1] \cap \mathbb{Q}$ to $\bar{\mathbb{R}}$ $$f(x) = \sum_{i = 1}^\infty 2^{-i}|x - x_i|^\alpha$$

I need to prove that $f$ is Lesbesque integrable.

If we define $f_n$ to be $\sum_{i = 1}^n2^{-i}|x - x_i|^\alpha$ we can conclude from the monotone convergence theorem that $\int f d\lambda = \lim_{n \to \infty} \int f_n d\lambda$.

If we can find a finite upper bound of $$\left\{ \int |x - x_i|\chi_Ad\lambda \ | \ i \in \mathbb{N}\right\} $$ where $A = [0, 1]$ we are basically done because $\sum_{i = 1}^\infty2^{-i}$ converges, but I don't know how to do this and if it is possible. We can take singularities out of $A$ if necessary because singular points are $\lambda$-null.

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  • $\begingroup$ You say $f$ is defined on $\mathbb Q \cap [0,1]$. What is Lebesgue measure on the rationals?. $\endgroup$ – Kavi Rama Murthy Nov 4 '18 at 13:11
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Just compute $\int_0^{1} |x-x_i|^{\alpha}$ explicitly by splitting the integral into $(0,x_i)$ and $(x_i,1)$. You will see that $\int_0^{1} |x-x_i|^{\alpha} =\frac 1 {\alpha+1| }(x_i^{1+\alpha}+(1-x_i)^{1+\alpha}) \leq \frac 2 {\alpha+1}$. Hence $\int f(x) dx\leq\sum\frac 1 {2^{n}}\frac 2 {\alpha+1} <\infty$. (You can always interchange sum and integral when the terms are non-negative).

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