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I am reading Swartz's book (Measure, Integration and function spaces) and I come across an example 12, p. 73 which kind of make sense to me but not really. This deals with the mapping in the complement of Cantor and flat-Cantor set.

I was wondering if someone has better example for my question.

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Sure. There are $\mathfrak c$ (continuum) many Borel subsets of $\mathbb R$, but there are $2^{\mathfrak c}$ many subsets of the standard Cantor set. Thus there must be a subset $A$ of the standard Cantor set which is not a Borel set. Take $g$ to be the indicator function of $A$, then $g$ is $0$ Lebesgue-a.e. and not Borel measurable.

  • 0
    How would you argue $0$ function is Borel measurable by the way, is it just trivially true? Or there is some standard way of proving it.2012-12-25
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    It's easy to prove that any continuous function is Borel measurable.2012-12-25
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    Oh I see, Thank you much.2012-12-25
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    But proof that a constant is Borel measurable is easier than proof that a general continuous function is Borel measurable.2012-12-25
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    @GEdgar, Can you show me how? Please!2012-12-28
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    A constant function $f$ has the property that for any real number $t$, the set $\{x : f(x) > t\}$ is either the empty set or the whole space. Therefore, $f$ is measurable.2012-12-28