Show that if $f$ is right continuous then $f$ is measurable

Question

I'm trying to solve the following problem:

Show that if the function $f: \mathbb{R} \to \mathbb{R}$ is right continuous then $f$ is measurable.

(A function $f$ is measurable if $f^{-1}(E) \in \mathcal{B}(\mathbb{R})$ for each $E \in \mathcal{B}(\mathbb{R})$ which is equivalent to say that $f^{-1}(\alpha, \infty) \in \mathcal{B}(\mathbb{R})$ or $f^{-1}[-\infty, \alpha] \in \mathcal{B}(\mathbb{R})$, $\forall \alpha \in \mathbb{R}$, where $\mathcal{B}(\mathbb{R})$ represents the Borel $\sigma$-algebra)

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I know that if $f$ is continuous then it is measurable since $(\alpha, \infty)$ is an open set and hence $f^{-1} (\alpha, \infty)$ would be open and therefore measurable with respect the Borel $\sigma$-algebra, but I'm not sure how could I construct a similar argument with right continuity.

Answer 1

Consider $f_n(x):=\sum_{k\in\Bbb Z} f(k/n)\cdot 1_{((k-1)/n,k/n]}$, $\in \Bbb R$. Each $f_n$ is Borel measurable, being a step function. And $\lim_nf_n(x)=f(x)$ for all $x\in\Bbb R$ because $f$ is right continuous. (To see this, fix $x$. Given $\epsilon>0$ choose $\delta>0$ so small that $|f(t)-f(x)|<\epsilon$ if $x\le t<x+\delta$. Now choose $n$ so large that $1/n<\delta$. There is a unique integer $k$ such that $(k-1)/n<x\le k/n$, and for that $n$ we have $|x-k/n|\le 1/n<\delta$, so $|f(x)-f(k/n)|<\epsilon$. It follows that $|f_n(x)-f(x)|<\epsilon$.)