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I'm trying to solve the following exercise:

Let $f,g:\mathbb R_+\rightarrow \mathbb R_+$ be given by $f(x) = \sum_{n=0}^\infty n\mathbb 1_{[2n,2n+2)}(x)$ and $g(x) = \sum_{n=0}^\infty\mathbb 1_{[2n,\infty)}(x)$. Is it true that: a) $\sigma(f) \subset \sigma(g)$ or b) $\sigma(g) \subset \sigma(f)$?

I tried to solve it this way:

$f^{-1}([0,a]) = [0, 2b+2)$, where $b \in \mathbb N_0, b = \max\{n\in \mathbb N_0: n \le a\}, a \in \mathbb R_+\setminus\{0\} $. Therefore $\sigma(f) = \sigma(\{[0, 2b+2): b \in \mathbb N_0\})$, since $\mathcal B(\mathbb R_+) = \sigma(\{[0,a]: a \in \mathbb R_+\setminus\{0\}\}).$

In turn, $g^{-1}([1,a]) = [0, 2c), c \in \mathbb N, c = \max\{n \in \mathbb N:n \le a\}, a \in \mathbb R_+, a \gt 1$. Therefore $\sigma(g) = \sigma(\{[0,2c): c \in \mathbb N$}).

So I obtain $\sigma(f) = \sigma(g)$. Do I miss something?

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1 Answer

$${}{}{}{}{}g=f+1{}{}{}{}{}{}$$

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Thanks, but I still can't find a mistake in my solution as well as how to use your observation to solve the exercise. –  Victor Oct 15 '12 at 20:04
    
You do not know how to use the fact that $g=f+1$ to prove that $\sigma(g)=\sigma(f)$? Really? Let $B\subset\mathbb N$. Let us show that $g^{-1}(B)$ is in $\sigma(f)$. For this, it suffices to show that... –  Did Oct 15 '12 at 20:15
    
$g^{-1}(B) = f^{-1}(C) \in \sigma(f)$, where C = $\{b-1: b \in B\}?$ –  Victor Oct 15 '12 at 21:00
    
What do you think? –  Did Oct 15 '12 at 21:07
    
I think it's correct, otherwise I wouldn't post the comment. (Also need to show that $\sigma(f) \subset \sigma(g)$, but it's the same). –  Victor Oct 15 '12 at 21:15
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