I am just a student in probability theory and I am trying to understand by myself why: $\int_{0}^{1}I_{\omega \notin \mathbb{Q}}(w) d\mu(\omega )=1$ with $\mu (\omega )$ the Lebesgues measure.
1-To show it i thought first: $\int_{0}^{1}I_{\omega \notin \mathbb{Q}}(\omega) d \mu (\omega)=\mu(\omega \notin \mathbb{Q})=1-\mu(\omega \in \mathbb{Q})$
2-So now i will focus on trying to proove that $\mu(\omega \in \mathbb{Q})=0$
To do it I ll proceed as follow.
First let pick an arbitrary rationnal number $x_1$ on $]0;1[$ that means $x_1 \in \mathbb{Q} \cap ]0;1[$
We know that it exists $(\alpha _1;\beta _1) \in \mathbb{Q}\cap [0;1]$ s.t. $\alpha _1< x_1<\beta _1$ So of course: $0 \leq \mathbb{P} (X=x_1) < \mathbb{P} (\alpha _1 \leq X \leq \beta _1)\leq \frac{1}{1}$ because $x_1 \in ]\alpha _1;\beta _1[$
Now we choose an other rationnal number $ x_1 \neq x_2 \in \mathbb{Q} \cap ]0;1[$ and we proceed exactly in the same way as for $x_1$ but now we require $(\alpha _2;\beta _2) \in \mathbb{Q}\cap [0;1]$ such that: $0 \leq \mu (X=x_2) < \mu (\alpha _2 \leq X \leq \beta _2)\leq \frac{1}{2^2}$. Such $(\alpha _2;\beta _2)$ of course exist (we can even construc them from $x_2$)
...etc...
We proceed exactly the same way for each rationnal number that we writte $x_n$ (that we know are countable). We know that for each of those rationnal number $x_n, \exists (\alpha _n;\beta _n)\in \mathbb{Q}\cap [0;1] \ \ s.t. \ \ \alpha _n<x_n<\beta _n$ and that verify:$0 \leq \mu (X=x_n) < \mu (\alpha _n \leq X \leq \beta _n)\leq \frac{1}{n^2}$
Rem: Such $\alpha _n,\beta _n$ always exist because for exemple we can construct them as follow: $\alpha _n=x_n-1/2n^3$ and $\beta _n=x_n+1/2n^3$
3-Now we have: $\sum_{0 \leq x_n \in \mathbb{Q} \leq 1}^{} \mu(X=x_n) \leq \sum_{0 \leq x_n \in \mathbb{Q} \leq 1}^{} \mu(\alpha_n \leq X \leq \beta_n) \leq \sum_{n=1}^{\infty } \frac{1}{n^2}<\infty
$
And by using BC lemma we cannow conclude that: $\mu(\omega \in \mathbb{Q})=0$
Q.E.D
Is this correct? I haven't somebody to really ask him this question so i will be happy to read your comment.
Thank you