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Consider the function $f(x)=\begin{cases} x & x\in \mathbb{Q}\cap [0,1] \\ -x & x\in [0,1]-\mathbb{Q} \\ \end{cases}$

How do you go about computing the following Lebesgue integral?

$$\int_0^{1}f(x)d\mu$$

I need to solve a similar problem, and though I found this on a different post from a year ago, the post dealt only with integrability and not how to solve it. My textbook does not have anything similar. Any help would be greatly appreciated.

user76286
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1 Answers1

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  • Prove the integral is defined (not difficult)
  • Write $\int_{[0,1]} fd\mu = \int_{[0,1]\cap\mathbb{Q}} fd\mu + \int_{[0,1]\setminus\mathbb{Q}} fd\mu$, and observe that
    • The set $[0,1]\cap\mathbb{Q}$ has measure $\mu( [0,1]\cap\mathbb{Q}) = 0$; what can you say about the first integral?
    • Similarly, as $\mu \mathbb{Q} = 0$, what can you say about $\int_{[0,1]} (-x)\mu(dx)$? Use it for the second one, as $$\int_{[0,1]\setminus\mathbb{Q}} f d\mu = \int_{[0,1]} (-x) \mu(dx) - \int_{\mathbb{Q}} (-x) \mu(dx)$$
Clement C.
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  • I am fairly new to this material, but by my limited understanding I think knowing that $\mu( [0,1]\cap\mathbb{Q}) = 0$ means that the first integral is equal to zero. – user76286 May 07 '13 at 15:04
  • As for the second part, if similar reasoning is applied, I am not sure why $\mu \mathbb{Q} = 0$, would give us an indication of $\int_{[0,1]} (-x)\mu(dx)$. Why does it not imply that $\int_{\mathbb{Q}}(-x)\mu(dx)$ is zero? Then the integral would be reduced to $\int_{[0,1]}(-x)\mu(dx)$ right? Or am I completely off? – user76286 May 07 '13 at 15:07
  • @user76286 Indeed, the first one is zero. And your second observation is also true (I might have phrase my last bullet in a slightly weird way); basically, the zero measure of any countable set gives you that $\int_{[0,1]\setminus\mathbb{Q}}(-x)\mu(dx) = \int_{[0,1]}(-x)\mu(dx)$; and this one is straightforwardness to compute. – Clement C. May 07 '13 at 15:22
  • Great! Then I do understand. Thanks so much for your help. – user76286 May 07 '13 at 15:28