The Coolest Proof

July 9, 2023

I was trying to solve Problem 5 from Chapter 1 of Rudin's Rudin and Complex Analysis. The problem is as follows: consider measurable functions $f$ and $g$ from some measurable space to the extended real numbers. Show that the set $\{x : f(x) < g(x)\}$ is measurable.

My first proof came from considering the difference $f - g$ which is measurable, and then using the definition of a measurable function which gives that the preimage of any open set is a measurable set. However, this proof wasn't super satisfying to me since it basically relied on the previous theorem that $f - g$ is measurable, and doesn't rely only on the definitions of sigma algebras and measurable spaces (and the respective continuous/measurable functions). Overall, for a while I held onto this strong inkling that there must be a better, more set-union-intersectiony topology-esque proof. But I thought about it, and for some reason I couldn't find it.

Today I was talking to my dad about it, and I finally suddenly discovered this proof: notice that $$\{x : f(x) < g(x)\} = \bigcup_{r \in \mathbb{Q}} (f^{-1}(A) \cap g^{-1}(B))$$ where $A$ is the interval $[-\infty,r)$ and $B$ in the interval $(r,\infty]$. $f^{-1}$ and $g^{-1}$ denote getting preimages. Since $A$ and $B$ are open, their preimages are measurable, and the rest follows from from the properties of sigma algebras.

I really like this proof a lot more because it doesn't use anything about the field structure of the real numbers, and only uses stuff about the topology of $\mathbb{R}$. Overall, I was also really proud of myself for coming up with this type of set-based topology-esque proof, since I haven't really made my own before and generally always just read them in textbooks.