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

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 ?

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 $B\in\mathcal{F}$.
• 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 ? – Shuzheng Sep 19 '14 at 17:40
• $\sigma^{-1}(\mathcal{F})$ 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 $B\in\mathcal{F}$. – Stefan Hansen Sep 19 '14 at 17:42
• 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 $\mathcal E \setminus \sigma^{-1}(\mathcal F) = \emptyset$ – Shuzheng Sep 19 '14 at 17:51
• You have defined $\mathcal{E}=\sigma^{-1}(\mathcal{F})$ in your post. – Stefan Hansen Sep 19 '14 at 18:04
• I'm sorry, I was confused. I forgot $\mathcal E := \sigma^{-1}(\mathcal F)$. – Shuzheng Sep 19 '14 at 18:04