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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?

1 Answers 1

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

  • 0
    Thanks, but I still can't find a mistake in my solution as well as how to use your observation to solve the exercise.2012-10-15
  • 0
    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...2012-10-15
  • 0
    $g^{-1}(B) = f^{-1}(C) \in \sigma(f)$, where C = $\{b-1: b \in B\}?$2012-10-15
  • 0
    What do you think?2012-10-15
  • 0
    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).2012-10-15
  • 0
    Perfect. More generally, if $g=\Psi(f)$ then $\sigma(g)\subseteq\sigma(f)$. Here $g=\Psi(f)$ and $f=\Phi(g)$ hence $\sigma(g)\subseteq\sigma(f)\subseteq\sigma(g)$.2012-10-15
  • 0
    Thank you. But the original solution is still correct? I mean, I'm rather interested in logic behind it, as it can be used to solve different problems where it's not possible to find a short solution.2012-10-16