While studying concepts of measurable functions, there was a theorem suggesting that if any measurable...

abbracciopj

abbracciopj

Answered

2022-06-29

While studying concepts of measurable functions, there was a theorem suggesting that if any measurable function is altered on a null set, its measurability still remains.
If f : E R is measurable, E M , g : E R is such that the set { x : f ( x ) g ( x ) } is null, then g is measurable.
They showed the following proof.
Consider the difference d ( x ) = g ( x ) f ( x ). It is zero except on a null set, so
{ x : d ( x ) > a } = { a null set a 0 a full set a < 0
Here, a full set is the complement of a null set. Since both null and full sets are measurable, d is a measurable function. g = f + d is thus measurable.
Now, I was curious whether the statement remains true if I change the set { x : f ( x ) g ( x ) } into { x : f ( x ) = g ( x ) } , i.e., differ at points in a full set, because there was no doubt if I alter the proof as the following.
The difference d ( x ) is still measurable since
{ x : d ( x ) > a } = { a full set a 0 a full set a < 0
However, the later statement is actually not plausible at all. Is there any contradictory logic among here?
Appreicate as always.

Answer & Explanation

robegarj

robegarj

Expert

2022-06-30Added 24 answers

Let f = 0 , A E be a non-measurable subset and N is a null set with N A = .
Then we let g ( x ) = { 1 , x A 0 , x N 1 , x E ( A N )
Thus, { x : d ( x ) > 0 } = A which is non-measurable. So here d ( x ) is not a measurable function.

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