Composition of inverse of measurable functions

Question

Given a random variable $X:(\Omega,A)\rightarrow (E,\Sigma)$ and a measurable function $f$, how do I get: $$ f(X)^{-1}(\Sigma)=X^{-1}(f^{-1}(\Sigma))\subset X^{-1}(\Sigma) $$

Answer 1

I preassume that $f$ is a function $E\to E$.

For every $B\subseteq E$ and every $\omega\in\Omega$ the following statements are equivalent:

• $\omega\in(f\circ X)^{-1}(B)$
• $(f\circ X)(\omega)\in B$
• $f(X(\omega))\in B$
• $X(\omega)\in f^{-1}(B)$
• $\omega\in X^{-1}(f^{-1}(B))$

And looking at first and last bullet we conclude that for every $B\subseteq E$ we have:$$(f\circ X)^{-1}(B)=X^{-1}(f^{-1}(B))$$

This leads directly to:$$(f\circ X)^{-1}(\Sigma)=X^{-1}(f^{-1}(\Sigma))\tag1$$

where: $$(f\circ X)^{-1}(\Sigma):=\{(f\circ X)^{-1}(B)\mid B\in\Sigma\}$$ and: $$X^{-1}(f^{-1}(\Sigma)):=\{X^{-1}(f^{-1}(B))\mid B\in\Sigma\}$$

Now note that measurability of $f$ means exactly that $f^{-1}(\Sigma)\subseteq\Sigma$ so that $(1)$ can be expanded to:$$(f\circ X)^{-1}(\Sigma)=X^{-1}(f^{-1}(\Sigma))\subseteq X^{-1}(\Sigma)\tag2$$