[proofplan] One direction is immediate from measurability because half-lines are Borel. For the converse, collect all Borel sets whose inverse images are measurable and show this sigma-algebra contains the half-lines that generate $\mathcal B(\mathbb R)$. [/proofplan] [step:Measurability implies the half-line condition] If $X$ is measurable, then for each $a\in\mathbb R$ the set $(-\infty,a]$ is Borel, so \begin{align*} \{\omega:X(\omega)\le a\}=X^{-1}(( -\infty,a])\in\mathcal F. \end{align*} [/step] [step:The half-line condition implies measurability] Assume $X^{-1}(( -\infty,a])\in\mathcal F$ for every $a\in\mathbb R$. Let \begin{align*} \mathcal G=\{B\in\mathcal B(\mathbb R):X^{-1}(B)\in\mathcal F\}. \end{align*} The class $\mathcal G$ is a sigma-algebra on $\mathbb R$ because inverse images commute with complements and countable unions. By assumption, $(-\infty,a]\in\mathcal G$ for every $a$. The Borel sigma-algebra $\mathcal B(\mathbb R)$ is generated by these half-lines, so $\mathcal G=\mathcal B(\mathbb R)$. Hence $X^{-1}(B)\in\mathcal F$ for every Borel set $B$, which is measurability of $X$. [/step]