# Proving a function is Lesbesque integrable

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.

• You say $f$ is defined on $\mathbb Q \cap [0,1]$. What is Lebesgue measure on the rationals?. – Kavi Rama Murthy Nov 4 '18 at 13:11

## 1 Answer

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).