Show that $f:\mathbb{R}\to\mathbb{R}$ is Borel measurable if $f$ is either right continuous or left continuous.

$f$ is Borel measurable if $\forall c\in\mathbb{R}$ the set $\{w\in\mathbb{R}:f(w) < c\}\in \mathcal{B}$

Suppose the function is left continuous. Then we have that

$$\lim_{x \nearrow a} f(x) = f(a)$$

From this we get that the set $\{w\in\mathbb{R}:f(w) < c\}$ is in fact a union of intervals in $\mathbb{R}$ and therefore in $\mathcal{B}$. Same argument can be used for right continuous functions if you use the set $\{w\in\mathbb{R}:f(w) > c\}$ instead.

Does this make sense? I'm not sure if it is rigorous enough, especially the part where I say it's a union of internals. Maybe I could construct that somehow?

• There is nothing wrong with what you have, although your 'definition' of Borel measurability is not the definition I would give. I would say that your definition depends on a lemma. The normal definition would be that the preimage of a Borel set is a Borel set. Your definition is equivalent, but prima facie weaker. Commented Oct 27, 2017 at 15:22
• @KyleFerendo Yes, I know. In practice it I never see proofs using the actual definition. I think it's just harder to think in terms of preimages. At least for me it is. Commented Oct 27, 2017 at 15:27
• Can you perhaps justify further why you feel that left-continuity immediately implies that the set in question is a union of intervals? That, I feel, is really the heart of the question. Commented Oct 27, 2017 at 15:29
• @SamT For any singleton in the set, $\{\dots\}$ you can use continuity to build an interval to one side or another of that singleton that is also in the set $\{\dots\}$. Hence you can construct $\{\dots\}$ from a union of intervals. Commented Oct 27, 2017 at 15:38
• Ok yeah. You said you were unsure on the rigour of that part, but if you were to write what you've just written, then I think that would be fine (notwithstanding KyleFerendo's comment) Commented Oct 27, 2017 at 17:50

Let $f$ be a left continuous function and let $O \subseteq \mathbb{R}$ be an open set. Let $E=f^{-1}(O)$. By definition, given $x \in E$ there exists a $\delta>0$ such that $(x-\delta ,x] \subseteq E$. Hence $E$ is a contable union of semi-open sets and therefore $E$ is Borel measurable.
If $f$ is right continuous then for every $y \in E$, there exists a $\delta’>0$ such that $[y,y+\delta’) \subseteq E$ and the argument is analogous.
• Why is $E$ a countable union of half open sets? Could you write an explicit countable union that verifies that? Commented Oct 11, 2018 at 19:29
• @Algebear I’m possibly a bit late, but I can answer that. So we have established that $E$ is a (probably uncountable) Union of half open intervals $[a,b)$. You can show that this always reduces to a countable Union of such sets as follows: Define an equivalence relation on $E$ by $x \sim y$ iff for all $z$ with $x<z<y$ one has $z \in E$, ie, $x$ and $y$ lie in the same interval. Then I believe you can show that the equivalence classes are intervals (and of course disjoint!). Conclude countability by taking a rational in each of these intervals. Commented Jan 22 at 8:26