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I need to show that an increasing function $f:\mathbb{R}\rightarrow \mathbb{R}$ is (borel)-mesurable

proof From the lecture we know that since $B(\mathbb{R})=\langle Q(\mathbb{R})^\sigma\rangle$, it's enough to show that $\forall E\in Q(\mathbb{R})\,\,\,f^{-1}(E)\in B(\mathbb{R})$. Therefore let $E$ be as above, then $E=\bigcup_{i=1}^n[a_i,b_i),\,\,\,s.t.a_1<b_1<a_2<b_2...$ then $$f^{-1}(E)=\bigcup_{i=1}^nf^{-1}([a_i,b_i))$$ Now we only need to show that $O_i=f^{-1}([a_i,b_i))$ are in $Q(\mathbb{R})$, and then by the characterisation of $B(\mathbb{R})$ we can conclude that $\bigcup_{i=1}^n O_i\in B(\mathbb{R})$.

To do so I only need to prove that the preimage of an intervall is again an intervall.

Thank you very much.

user123234
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  • How does disjointness come into the picture? Union of any sequence of sets in $B(\mathbb R)$ belongs to $B(\mathbb R)$. – Kavi Rama Murthy Oct 17 '21 at 08:11
  • sorry I don't understand your point. So I want to show that $\bigcup_{i=1}^n f^{-1}([a_i,b_i))\in B(\mathbb{R})$. But since $B(\mathbb{R})$ is a $\sigma$-algebra we know that if $f^{-1}([a_i,b_i))\in B(\mathbb{R})$ and they are disjoint, then also their union is in $B(\mathbb{R})$. Therefore I wanted to show this two points. – user123234 Oct 17 '21 at 08:29
  • Could you maybe help me if my Idea is totally wrong? – user123234 Oct 17 '21 at 09:05
  • $\langle Q(\mathbb{R})^\sigma\rangle$ is not a satndar notation and I really don't know what you are doing. But the usual proof of this result is very simple. You only need the fact that inverse image of any interval is an interval. – Kavi Rama Murthy Oct 17 '21 at 09:21
  • Sorry I remarked that I hat an error in my definition of a sigma-algebra, because there it was that the union of disjoint sets in the sigma algebra is again in the sigma algebra, but I saw now that these sets don't have to be disjoint. That's why I wanted to prove this fact. – user123234 Oct 17 '21 at 09:40
  • So I understand now why it's enough to show that the preimage of any interval is a interval, but how can i prove this since f is not necessairly continuous – user123234 Oct 17 '21 at 09:47
  • An interval is a set $I$ such that $x,y \in I, x<y$ implies every point $z $ with $x<z<y$ is also in $I$. Use this to show that the inverse image of any interval is also an interval. Continuity is not required for this. – Kavi Rama Murthy Oct 17 '21 at 09:55
  • don't you mean that z is in I if $x<z<y$? – user123234 Oct 17 '21 at 09:58
  • So can I prove it like this: Let $I$ be an interval. We want to show that $f^{-1}(I)$ is also an interval. Let $x,y \in f^{-1}(I)$ such that $x<y$. This is equivalent to say that $f(x),f(y)\in I$, but since f is increasing we get that $f(x)<f(y)$. By the definition of an interval we knot that for each $f(z)$ s.t. $f(x)<f(z)<f(y)$ we have that $f(z)\in I\Leftrightarrow z\in f^{-1}(I)$. – user123234 Oct 17 '21 at 10:12

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