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

Question

$f$ 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.

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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?

Answer 1

Here is an easier proof (in my opinion).

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.