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I have this exercise:

Suppose $f$ is a real valued continuous function on $[a,b]$. Show that $f(B)$ is a Borel set for every Borel subset $B$ of $[a,b]$.

Hint: Consider the collection $M$ of all subsets $A$ of $[a,b]$ for which $f(A)$ is a Borel set. Show that $M$ is a sigma-algebra.

I am struggling a little. The hint tells me to look at:

$M=\{A\mid f(A) \text{ is Borel}\}$, and show that this is a sigma algebra. I do get that $\emptyset \in M$, and that M is closed under countable unions. But how do I get closed under complements?, I mean $A \in M \rightarrow A^c \in M$?, the problem is that we only have $f(A)^c=f(A^c)$ if f is bijective.

And to finish the proof I also need that $f(O)$ is borel for every open set, but how do I get this?

user119615
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    The image of a Borel set under a continuous map is, in general, not Borel. – saz Nov 06 '15 at 08:02
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    This is the mistake that Lebesgue, himself, did in his one of articles ! – Fardad Pouran Nov 06 '15 at 08:34
  • The hint is also wrong: since all closed subsets of $[a,b]$ have a Borel image, if $M$ were a $\sigma$-algebra, it would contain the $\sigma$-algebra generated by closed sets, which is the Borel $\sigma$-algebra. That's probably the idea of the hint. But, while you can prove that a countable union of elements of $M$ is in $M$, it's not enough to show it's a $\sigma$-algebra. Since we know there is a Borel set not in $M$, then $M$ can't be a $\sigma$-algebra. Actually, while there is a nice equality $f(\bigcup_n A_n)=\bigcup_n f(A_n)$, there is no such thing for the complement. – Jean-Claude Arbaut Nov 06 '15 at 10:18

1 Answers1

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The statement you want to prove is wrong: There is a Borel set $B \subseteq [0,1]^2$ such that its projection $\pi_1[B]$ onto the first coordinate is not Borel (see for example here). Now let $\gamma \colon [0,1] \to [0,1]^2$ a continuous onto map, and $f := \pi_1 \circ \gamma \colon [0,1]\to [0,1]$. Then, $f$ is continuous, $A := \gamma^{-1}[B]$ is Borel, but $$ f[A] = \pi_1[\gamma\gamma^{-1}[B]] = \pi_1[B] $$ is not.

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martini
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  • Did you try further? I do not know at what step of the Borel hierarchy the statement breaks down, I'm sure that is also an interesting question. – martini Nov 06 '15 at 08:33
  • So far, I have this: the continuous image of a compact is compact, and any closed subset of $[a,b]$ is compact, hence the image of a closed subset is a Borel set. Since $f(\bigcup_n A_n)=\bigcup_n f(A_n)$, taking $A_n$ to be closed, you get that the image of a $F_\sigma$ is also a Borel set. – Jean-Claude Arbaut Nov 06 '15 at 09:26
  • Some further reading lead me to the result, that the $B$ above can be choosen to be a $G_\delta$, so nothing beyond $F_\sigma$ will work. – martini Nov 06 '15 at 10:03