Let $f:\mathbb{R}_+ \rightarrow \mathbb{R}$ be a right-continuous function, i.e., $f(r_n) \rightarrow f(r)$ whenever $(r_n)$ is a sequence decreasing to $r$. Prove that $f$ is $\mathcal{B}$-measurable, i.e., $\forall{a} \in \mathbb{R}$, $f^{-1} (-\infty,a) \in \mathcal{B}_{\mathbb{R}}$.
My attempt
Recall that elementary functions are measurable, i.e. $f:\mathbb{R} \rightarrow \mathbb{R}$ is $\mathcal{B}$-measurable whenever it has the form $f=\sum_{n=1}^{\infty}a_n\mathbb{I}_{A_n}$.
Consider the function defined by $d_n(r)=\sum_{k=1}^{\infty} \frac{k}{2^n} \mathbb{I}_{\big[\frac{k-1}{2^n}, \frac{k}{2^n}\big)}$.
Note that, $\forall n \geq 1$, $d_n$ is $\mathcal{B}$-measurable as $\forall{a} \in \mathbb{R} \; d_n^{-1}(-\infty,a)=\bigcup_{k=1}^{2^{n}a}\big[ \frac{k-1}{2^n}, \frac{k}{2^n} \big)=(0,a) \in \mathcal{B}$.
Note that $\forall n \geq 1$, $f_n=f \circ d_n$ is of the form $f_n=\sum_{k=1}^{\infty}a_{k,n} \mathbb{I}_{A_{k,n}}$. Precisely,
$f_n(x)=f \big( \sum_{k=1}^{\infty} \frac{k}{2^n} \mathbb{I}_{\big[\frac{k-1}{2^n}, \frac{k}{2^n}\big)}(x) \big) = \sum_{k=1}^{\infty} f \big( \frac{k}{2^n}\big)\mathbb{I}_{\big[\frac{k-1}{2^n}, \frac{k}{2^n}\big)}(x)$,
where the last equality holds because the sets $\big[ \frac{k-1}{2^n}, \frac{k}{2^n} \big)$ are disjoint. So, $\forall n \geq 1 \; f_n$ is an elementary function, and as such it is $\mathcal{B}$-measurable. Also, $f_n \rightarrow f$ from above by right-continuity of $f$, which implies that $f$ is $\mathcal{B}$-measurable.
I don't know if this solution is correct, most of all I am afraid some steps might be incorrect or need some intermediate steps. Any help would be much appreciated, thank you!
The function $d_n(r)$ encounters an incorrect use of the index $n$. Probably what you meant to say $$d_n(r)=\sum_{k=1}^{2^n} \frac{k}{2^n} \mathbb{I}{\left [ \frac{k-1}{2^n}, \frac{k}{2^n} \right )}(r)$$
$d_n$ is Borel measurable because each of the half-intervals on which it is defined is a Borel set, and a countable union of Borel sets is also Borel. The definition of $f_n$ looks correct if $m_n = 2^n$. You use step functions that are Borel measurable and you approximate $f$ with a sequence of these functions. It was also correctly pointed out that $f_n \rightarrow f$ is from above because of the right-hand continuity of $f$, but it was better to justify this step, more explicitly. You must point out that for any $x \in \mathbb{R}_+$, due to the right continuity of $f$, $f_n(x)$ converges to $f(x)$ as $n \to \infty$
On the whole, the evidence is acceptable!