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
$\displaystyle A^*$ is a $\displaystyle \sigma$-$\displaystyle algebra$ on $\displaystyle \Omega^*$
it follows that
$\displaystyle {\Omega^*}\in{A^*}$
If $\displaystyle f^{-1}$ exists, it follows that $\displaystyle f$ is bijective.
Thus $\displaystyle f^{-1}(\Omega^*)=\Omega$
Therefore $\displaystyle \Omega\in{A}$.
Similarly, given that
$\displaystyle \forall{E^*}\in{A^*},{({E^*})^c}\in{A^*}$
it follows that
$\displaystyle f^{-1}{({E^*})^c}\in{A}$
Then you have left only to show:
(iii) $\displaystyle \forall$ infinite sets $\displaystyle {(A_i)}\in{A}, \cup{A_i}\in{A}$
(iv) $\displaystyle \forall$ infinite sets $\displaystyle {(A_i)}\in{A}, \cap{A_i}\in{A}$
Can I ask which book your university is using for Measure Theory?
TXGirl, the use of notation, $\displaystyle f^{ - 1}$, is not the inverse of a function.
It is confusing notation. That is exactly why many of us prefer a different notation.
$\displaystyle f:A \mapsto B\quad \wedge C \subseteq B\quad \Rightarrow \quad \overleftarrow f (C) = f^{ - 1} [C] = \left\{ {x \in A:f(x) \in C} \right\}$.
In other words, we now have a mapping $\displaystyle \overleftarrow f :P(B) \mapsto P(A)\,\& \,\overleftarrow f (B) = A$.
The second property comes from the way mappings are defined.
Ok, so the goal then seems to be to show that
$\displaystyle {A}$:=$\displaystyle \left\{ ({x}\in{\Omega}:f(x)\in{E^*}):{E^*}\in{A^*} \right\} $ is a $\displaystyle \sigma$-algebra on $\displaystyle \Omega$
So your first step is:
(i) Show $\displaystyle \Omega\in{A}$
You have that $\displaystyle f(\Omega)\subseteq {\Omega^*}\in{A^*}$ the latter inclusion following from the fact that $\displaystyle {A^*}$ is a $\displaystyle \sigma$-algebra on $\displaystyle {\Omega^*}$
Thus it follows that $\displaystyle {\Omega}\in{A}$
(ii) Show $\displaystyle \forall{X}\in{A}, {X^c}\in{A}$
If $\displaystyle X \in A \quad \Rightarrow \quad f(X)\in{A^*}$
$\displaystyle \quad \Rightarrow \quad {(f(X))^c}\in{A^*}$
$\displaystyle \quad \Rightarrow \quad {f(X^c)}\in{A^*}$
$\displaystyle \quad \Rightarrow \quad {X^c}\in{A}$
(iii) Show then that $\displaystyle \forall$ infinite sets $\displaystyle {(A_i)}\in{A}, \cup{A_i}\in{A}$
(iv) Finally show that $\displaystyle \forall$ infinite sets $\displaystyle {(A_i)}\in{A}, \cap{A_i}\in{A}$
**Note: $\displaystyle {X^c}$ denotes the complement of $\displaystyle {X}$ in the relevant space.
Here is the first part: $\displaystyle \Omega ^* \in A^* \Rightarrow \quad \Omega = \overleftarrow f \left( {\Omega ^* } \right) \in A$.
If we have $\displaystyle \overleftarrow f \left( {E^* } \right) \in A \Rightarrow \quad \Omega \backslash \overleftarrow f \left( {E^* } \right) = \overleftarrow f \left( {\Omega ^* \backslash E^* } \right) \in A$.
If we have a countable collection $\displaystyle \left\{ {\overleftarrow f \left( {E_\alpha ^* } \right)} \right\} \Rightarrow \quad \overleftarrow f \left( { \cup E_\alpha ^* } \right) = \cup \left( {\overleftarrow f \left( {E_\alpha ^* } \right)} \right) \in A$.