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I was doing some self-studying and came upon the following question.

Suppose $a < b$ and let $M([a,b])$ denote the Lebesgue measurable subsets of $[a,b]$. Define the function $f: [0,1] \rightarrow [a,b]$ by $f(x) = mx + a$ where $m = b-a$. Show that the correspondence $A \mapsto f(A)$ is a bijection from $M([0,1])$ to $M([a,b])$.

While trying to show surjectivity, I ran into the problem of not knowing how to show if a set is measurable in $[0,1]$. I was wondering how to approach this problem.

Thanks!

MPW
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guest
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  • Note that since $f$ is continuous, it is measurable. Hence, if $A$ is a measurable subset of $[a,b]$, $f^{-1}(A)$ is a measurable subset of $[0,1]$ by definition. – Brandon Jul 31 '14 at 16:11
  • Thanks for the answer. The book hasn't covered measurable functions yet so I am not sure if I am allowed to use that in the argument... – guest Jul 31 '14 at 16:17
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    @Brandon: Even better: $f$ is a homeomorphism, so both $f$ and $f^{-1}$ are continuous – MPW Jul 31 '14 at 16:17
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    Continuous functions need not be Lebesgue-Lebesgue measurable. – Jonas Dahlbæk Jul 31 '14 at 16:18
  • Continuous functions are Borel-Borel measurable. Being Lebesgue-Lebesgue measurable is another condition. See this – Jonas Dahlbæk Jul 31 '14 at 16:28

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Since conservation under continuous maps is not entirely trivial it is better to use the linearity of your map. Here just about any characterisation of measurability will work, for example equality of inner and outer measures. Both of the latter trandform by $b$ when you apply $f$ and therefore they coincide in the source iff they coincide in the target.

Mikhail Katz
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My favorite way to do this is by noting that the Lebesgue measure is a Haar measure on $\Bbb R$, and that it is sufficient to prove that the transformation preserves measurability on $\Bbb R$, since it's easy to see that measurable sets in the two intervals are just inherited from those on $\Bbb R$.

Consider the altered "measure" $E\mapsto \mu(mE)$, where $\mu$ is the Lebesgue measure. Then this candidate is also translation invariant, hence is also a Haar measure, i.e. is a multiple of the Lebesgue measure (with the multiple being $|m|$). Hence all translates of dilations are Lebesgue measurable, since multiplying a measure by a constant does not change the $\sigma$-algebra.

Adam Hughes
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