# $f$ Borel measurable and and $f=g$ a.e (Lebesgue) but $g$ is not Borel measurable

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.
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. –  Deepak Dec 25 '12 at 19:17
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. –  GEdgar Dec 28 '12 at 19:48