1. ## measure theory question

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?

2. Given that

$A^*$ is a $\sigma$- $algebra$ on $\Omega^*$

it follows that

${\Omega^*}\in{A^*}$

If $f^{-1}$ exists, it follows that $f$ is bijective.

Thus $f^{-1}(\Omega^*)=\Omega$

Therefore $\Omega\in{A}$.

Similarly, given that

$\forall{E^*}\in{A^*},{({E^*})^c}\in{A^*}$

it follows that

$f^{-1}{({E^*})^c}\in{A}$

Then you have left only to show:

(iii) $\forall$ infinite sets ${(A_i)}\in{A}, \cup{A_i}\in{A}$

(iv) $\forall$ infinite sets ${(A_i)}\in{A}, \cap{A_i}\in{A}$

Can I ask which book your university is using for Measure Theory?

3. TXGirl, the use of notation, $f^{ - 1}$, is not the inverse of a function.
It is confusing notation. That is exactly why many of us prefer a different notation.
$f:A \mapsto B\quad \wedge C \subseteq B\quad \Rightarrow \quad \overleftarrow f (C) = f^{ - 1} [C] = \left\{ {x \in A:f(x) \in C} \right\}$.
In other words, we now have a mapping $\overleftarrow f :P(B) \mapsto P(A)\,\& \,\overleftarrow f (B) = A$.
The second property comes from the way mappings are defined.

4. Originally Posted by Plato
TXGirl, the use of notation, $f^{ - 1}$, is not the inverse of a function.
It is confusing notation. That is exactly why many of us prefer a different notation.
$f:A \mapsto B\quad \wedge C \subseteq B\quad \Rightarrow \quad \overleftarrow f (C) = f^{ - 1} [C] = \left\{ {x \in A:f(x) \in C} \right\}$.
In other words, we now have a mapping $\overleftarrow f :P(B) \mapsto P(A)\,\& \,\overleftarrow f (B) = A$.
The second property comes from the way mappings are defined.

Ok, so the goal then seems to be to show that

${A}$:= $\left\{ ({x}\in{\Omega}:f(x)\in{E^*}):{E^*}\in{A^*} \right\}$ is a $\sigma$-algebra on $\Omega$

(i) Show $\Omega\in{A}$

You have that $f(\Omega)\subseteq {\Omega^*}\in{A^*}$ the latter inclusion following from the fact that ${A^*}$ is a $\sigma$-algebra on ${\Omega^*}$

Thus it follows that ${\Omega}\in{A}$

(ii) Show $\forall{X}\in{A}, {X^c}\in{A}$

If $X \in A \quad \Rightarrow \quad f(X)\in{A^*}$
$\quad \Rightarrow \quad {(f(X))^c}\in{A^*}$
$\quad \Rightarrow \quad {f(X^c)}\in{A^*}$
$\quad \Rightarrow \quad {X^c}\in{A}$

(iii) Show then that $\forall$ infinite sets ${(A_i)}\in{A}, \cup{A_i}\in{A}$

(iv) Finally show that $\forall$ infinite sets ${(A_i)}\in{A}, \cap{A_i}\in{A}$

**Note: ${X^c}$ denotes the complement of ${X}$ in the relevant space.

5. Here is the first part: $\Omega ^* \in A^* \Rightarrow \quad \Omega = \overleftarrow f \left( {\Omega ^* } \right) \in A$.

If we have $\overleftarrow f \left( {E^* } \right) \in A \Rightarrow \quad \Omega \backslash \overleftarrow f \left( {E^* } \right) = \overleftarrow f \left( {\Omega ^* \backslash E^* } \right) \in A$.

If we have a countable collection $\left\{ {\overleftarrow f \left( {E_\alpha ^* } \right)} \right\} \Rightarrow \quad \overleftarrow f \left( { \cup E_\alpha ^* } \right) = \cup \left( {\overleftarrow f \left( {E_\alpha ^* } \right)} \right) \in A$.

6. one question, f with a backward dash means f^(-1)?

7. ahh sorry just saw it. many thanks

8. Originally Posted by TXGirl
Ok, so the goal then seems to be to show that

${A}$:= $\left\{ ({x}\in{\Omega}:f(x)\in{E^*}):{E^*}\in{A^*} \right\}$ is a $\sigma$-algebra on $\Omega$

(i) Show $\Omega\in{A}$

You have that $f(\Omega)\subseteq {\Omega^*}\in{A^*}$ the latter inclusion following from the fact that ${A^*}$ is a $\sigma$-algebra on ${\Omega^*}$

Thus it follows that ${\Omega}\in{A}$

(ii) Show $\forall{X}\in{A}, {X^c}\in{A}$

If $X \in A \quad \Rightarrow \quad f(X)\in{A^*}$
$\quad \Rightarrow \quad {(f(X))^c}\in{A^*}$
$\quad \Rightarrow \quad {f(X^c)}\in{A^*}$
$\quad \Rightarrow \quad {X^c}\in{A}$

(iii) Show then that $\forall$ infinite sets ${(A_i)}\in{A}, \cup{A_i}\in{A}$

(iv) Finally show that $\forall$ infinite sets ${(A_i)}\in{A}, \cap{A_i}\in{A}$

**Note: ${X^c}$ denotes the complement of ${X}$ in the relevant space.