Half open intervals in a Borel $\sigma$-algebra

Question

I am working on the exercise: prove that a right continuous function $\mathbb{R} \to \mathbb{R}$ is Borel measurable. I found that for every $x \in f^{-1}((a,b))$ we must have $$ [x, x + \delta_x) \subset f^{-1}((a,b)), $$ for some $\delta_x > 0$ depending on $x$. Consequently $$ \{ [x, x + \delta_x) \mid x \in f^{-1}((a,b)) \} = f^{-1}((a,b)). $$ I know that intervals are Borel measurable, but to prove that $f^{-1}((a,b))$ is measurable I would have to prove that $f^{-1}((a,b))$ is a countable union of such intervals. Could anybody give me a hint on how to do this?

Answer 1

Answer #2, corrected.

look at your sets $[x,\delta _x)$. Now, the claim is we can partition $f^{-1}(a,b)$ into maximal lengths of such intervals.

So, take a look at an arbitrary point $y\in f^{-1}(a,b)$.
It's in some arbitrary amounts of sets of the form $[x^*,\delta _{x^*})$, for $x^{*}\in(a,b)$. Let ${c_y}$ be the infinum of all of the $x^*$'s, let ${d_y}$ be the supremum of all of the $\delta _{x^*}$'s that y is a member of. Let these inf, sup be from the extended reals, so they are guaranteed to exist.

Claim: $\{[c_y,d_y):y\in f^{-1}(a,b)\}$ is a partition of $f^{-1}(a,b)$.

Proof of claim: everything in our set is clearly in one of these intervals, and nothing from outside of our set is in any of these intervals, so the union is our set in question. The only thing to prove is that they are disjoint...and I'm going to bed now, alas. if this still needs more work, I'll get to it tonight :)