I had this little thought bothering me:
Let be a finite measure space , two functions , and a sub - algebra.
If a.e. and is -measurable,
is -measurable too?
Im thinking yes but im not sure how to prove it.
Any ideas?
thnx
You reversed the roles of f and g but it is ok, lets do it your way .
i was considering that decomposition of the inverse mapping but got stuck in the following part.
Lets call the exceptional set where f and g differ , and
First question:
The left side of is
which is obviously in but i am not sure why it must be in since
The right side
... and thus
but i dont know what to do next..
thank you southprkfan1
Well it turns out that even when considering only one sigma algebra like you did the statement may not be true if the measure space is not complete in the sense that
is complete if the all the null sets are measurable.
So well, lets add the " complete" to the hypothesis.
The case when we have only one sigma algebra is true.
The other one remains yet unclear.
Well the completeness is just a technical aspect that most of the time is overlooked because if you have a measure over which is not complete you can always extend it to a measure over a bigger sigma algebra
satisfying that
This is done using Carathedory's Extension theorem. In fact if the initial measure is finite then the extension is unique.