# Proof of a theorem on measurable functions

Suppose $(\Omega, \mathscr{L})$ and $(S, \mathscr{B})$ are measure spaces and that a collection of sets $\mathscr{A}$ generates $\mathscr{B}$, i.e., $\sigma(\mathscr{A}) = \mathscr{B}$. Let $X: \Omega \rightarrow S$. If $X^{-1}(A) \in \mathscr{L} \quad \forall \space A \in \mathscr{A}$, then $X$ is measurable $\mathscr{B}$ (or $X$ is a random variable).

• Thanks for taking our remarks on your previous question into account. Commented Sep 15, 2016 at 9:08
• @justt I am sorry for that behaviour. Commented Sep 15, 2016 at 9:12

The collection of subsets of $S$ that have a preimage under $X$ in $\sigma$-algebra $\mathcal L$ can be shown to be a $\sigma$-algebra.

This is not really difficult to prove, since preimages are very coöperative in this.

So if this collection contains $\mathcal A$ as a subcollection then it will also contain $\mathcal B=\mathcal\sigma(\mathcal A)$ as a subcollection.

That means exactly that $X$ is measurable.

Let $\mathcal M = \{B\in \mathscr B : X^{-1}(B) \in \mathscr L\}$. You want to show that $\mathcal M$ is a monotone class and use the monotone class theorem, which states that if $M(\mathscr A)$is the smallest monotone class containing the algebra $\mathscr A$, then $M(\mathscr A) = \sigma(\mathscr A) = \mathscr B$. Now since by hypothesis $\mathcal M$ contains $\mathscr A$, and itself is a monotone class, then $\mathcal M$ contains $M(\mathscr A) = \mathscr B$. So $\mathcal M = \mathscr B$ which is exactly saying that $X$ is measurable.

• The answer of @drhab is way simpler, the monotone class theorem is best suited for situations where it is hard to show that some collection is a sigma-algebra. Commented Sep 15, 2016 at 9:19
• Yes, just use the definition and show that $\mathcal M$ is stable by countable union and finite intersection ! Commented Sep 15, 2016 at 9:25
• @sv_jan5 It all boils down to the properties $f^{-1}(\bigcup_{i\in I} A_i) = \bigcup_{i\in I} f^{-1}(A_i)$, and $f^{-1}(\bigcap_{i\in I} A_i) = \bigcap_{i\in I} f^{-1}(A_i)$, for any arbitrary (finite, countable, even uncountable) index set $I$. Same proof as before ! Remark that replacing preimages by images in the intersection property doesn't work : $f(A\cap B)$ might be smaller than $f(A) \cap f(B)$. That's why drhab highlighted that preimages are cooperative. Images are not ! Commented Sep 15, 2016 at 16:45
• Exactly as @justt says. Also $f^{-1}(A^c)=f^{-1}(A)^c$ deserves to be mentioned. Commented Sep 15, 2016 at 16:54
• $\dots\iff f(x)\notin A\iff x\notin f^{-1}(A)\iff x\in f^{-1}(A)^c$. Commented Sep 16, 2016 at 7:17