0
$\begingroup$

Factor theorem for $\bar {\mathcal M}(\mathcal E)^+$ (set of $\mathcal E$-$\mathcal B(\mathbb R)$-measurable functions with values in $[0,\infty]$).

Let $X$ be a non-empty set, let $(Y,\mathcal F)$ a measurable space and $\sigma: X \rightarrow Y$ be a function.

Then $\mathcal E := \sigma^{-1}(\mathcal F)$ is a $\sigma$-algebra in $X$, the smallest for which $\sigma$ is $\mathcal E$-$\mathcal F$-measurable.

I want to show $\bar {\mathcal M}(\mathcal E)^+ = \{ f \circ \sigma : f \in \bar {\mathcal M}(\mathcal F)^+ \}$.

I see that $\{ f \circ \sigma : f \in \bar {\mathcal M}(\mathcal F)^+ \} \subseteq \bar {\mathcal M}(\mathcal E)^+$, and I have a theorem saying the following (which I've proven):

If $V \subseteq \bar {\mathcal M}(\mathcal E)^+$ and $V$ satisfies:

(i) $1_A \in V$ for $A \in \mathcal E$

(ii) $f,g \in V$ and $\alpha, \beta \ge 0 \Rightarrow \alpha f + \beta g \in V$

(iii) $(f_n)$ is an increasing sequence of functions fra $V \Rightarrow \lim_{n \rightarrow \infty} f_n \in V$

then $V = \bar {\mathcal M}(\mathcal E)^+$.


I have had no trouble in proving (ii) and (iii). However, (i) is really causing my trouble.

A hint is given that $1_B \circ \sigma = 1_{\sigma^{-1}(B)}$ (how is this proved ?).

Even applying this hint, I've not yet solved (i) - could someone help me out ?

$\endgroup$
1
$\begingroup$

For $B\in\mathcal{F}$ we have that $\mathbf{1}_B\in\overline{\mathcal{M}}(\mathcal{F})^+$ and that $$ (\mathbf{1}_B\circ \sigma) (x)=1\iff\sigma(x)\in B\iff x\in \sigma^{-1}(B)\iff \mathbf{1}_{\sigma^{-1}(B)}(x)=1 $$ for all $x\in X$ showing that $\mathbf{1}_B\circ \sigma = \mathbf{1}_{\sigma^{-1}(B)}$. In particular, we have $\mathbf{1}_A\in V$ for all $A\in\mathcal{E}$ since such a set can be written as $\sigma^{-1}(B)$ for some $B\in\mathcal{F}$.

$\endgroup$
  • $\begingroup$ I thought of this too, but how do you know that there for every $A \in \mathcal E$ exist some $B \in \mathcal F$ such that $\sigma^{-1}(B) = A$ ? How can I see $\sigma^{-1}$ has this property ? I know that $\sigma^{-1}(B) \in \mathcal E$, but do I know it has this surjective property ? $\endgroup$ – Shuzheng Sep 19 '14 at 17:40
  • $\begingroup$ $\sigma^{-1}(\mathcal{F})$ is shorthand notation for $\{\sigma^{-1}(B)\mid B\in\mathcal{F}\}$, thus any set $A\in\sigma^{-1}(\mathcal{F})$ is of the form $\sigma^{-1}(B)$ for some $B\in\mathcal{F}$. $\endgroup$ – Stefan Hansen Sep 19 '14 at 17:42
  • $\begingroup$ Sorry, if I might be wrong. But then I can only conclude that $1_A \in V$ for every $A \in \sigma^{-1}(\mathcal F)$ ? What about $A \in \mathcal E \setminus \sigma^{-1}(\mathcal F)$ ? I see that I'm done if $\mathcal E \setminus \sigma^{-1}(\mathcal F) = \emptyset$ $\endgroup$ – Shuzheng Sep 19 '14 at 17:51
  • $\begingroup$ You have defined $\mathcal{E}=\sigma^{-1}(\mathcal{F})$ in your post. $\endgroup$ – Stefan Hansen Sep 19 '14 at 18:04
  • $\begingroup$ I'm sorry, I was confused. I forgot $\mathcal E := \sigma^{-1}(\mathcal F)$. $\endgroup$ – Shuzheng Sep 19 '14 at 18:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.