A* is a σ-algebra on a set Ω*, and let f be a function from a set Ω to Ω*. prove that A={f^(-1)(E*):E* is an element of A*} is a σ-algebra on Ω. (σ-algebra is said to be generated by f).
i'm not sure how to do this question, can somebody help me?
A* is a σ-algebra on a set Ω*, and let f be a function from a set Ω to Ω*. prove that A={f^(-1)(E*):E* is an element of A*} is a σ-algebra on Ω. (σ-algebra is said to be generated by f).
i'm not sure how to do this question, can somebody help me?
Given that
is a - on
it follows that
If exists, it follows that is bijective.
Thus
Therefore .
Similarly, given that
it follows that
Then you have left only to show:
(iii) infinite sets
(iv) infinite sets
Can I ask which book your university is using for Measure Theory?
TXGirl, the use of notation, , is not the inverse of a function.
It is confusing notation. That is exactly why many of us prefer a different notation.
.
In other words, we now have a mapping .
The second property comes from the way mappings are defined.
Ok, so the goal then seems to be to show that
:= is a -algebra on
So your first step is:
(i) Show
You have that the latter inclusion following from the fact that is a -algebra on
Thus it follows that
(ii) Show
If
(iii) Show then that infinite sets
(iv) Finally show that infinite sets
**Note: denotes the complement of in the relevant space.