I was asked during class to show the following statment:
If $f$ is a measurable function and $g = f$ almost everywhere, then $g$ is measurable.
This is simple to show under the Lebesgue measure (or any complete measure) but is this true in general? If so, how would one go about showing it?
Suppose $K$ is the Cantor set. Then $K$ is Borel measurable, with $m(K) = 0.$ It is well knows that the cardinality of the set of all Borel subsets of $\mathbb {R}$ is $c,$ the cardinality of $\mathbb {R}.$ Recall $K$ also has cardinality $c,$ hence its set of subsets has cardinality $2^c > c.$ Thus there must exist subsets $E$ of $K$ that are not Borel sets. Fix any one of these. Then define $g=1$ on $E, g = 2$ on $K\setminus E,$ and $g=0$ everywhere else. Then $g$ is not Borel measurable, but it equals a Borel measurable function (namely $0$) on $\mathbb {R}\setminus K.$