# f is differentiable implies f' is borel, is this proof correct?

The usual proof (Differentiable function has measurable derivative?) makes use of the fact that a sequence of borel measurable functions converges to a borel mesurable function. However, the idea to use this property would have never crossed my mind so I solved it as follows, please point out whether this proof is correct or not :

Assume $$f$$ is differentiable. We want to prove that $$f'^{-1}(- ∞ ,a)$$ is Borel $$\forall a \in \mathbb{R}$$

i.e $$\{x\in \mathbb{R}$$ s.t $$f'(x) is Borel.

We know that $$f implies $$F where $$F$$ (resp. $$G$$) is the antiderivative of $$f$$ (resp. $$g$$)

using this fact : $$f'^{-1}(- ∞ ,a)= \{x\in \mathbb{R}$$ s.t $$f'(x) s.t$$f(x) where $$a$$ and $$b \in\mathbb{R}$$ (antiderivative of $$a$$ w.r.t $$x$$)

$$f'^{-1}(- ∞ ,a)= \{x\in \mathbb{R}$$ s.t $$f(x)-ax where : $$g(x) = f(x) - ax$$; $$a$$ and $$b$$ both being arbitrary real numbers. since $$g$$ is borel measurable (sum of $$f$$ and $$-ax$$ both borel measurable) therefore $$g^{-1}(- ∞ ,b) =f'^{-1}(- ∞ ,a)$$ is Borel $$\forall a \in \mathbb{R}$$

Any help would be appreciated

• In general $f < g$ does not imply that $F < G$. Consider $f = 0$, $g=2$, then one may take $F = 0$, $G = 2x$, and clearly $F \not < G$. Commented Oct 24, 2022 at 13:51
• $\{x\in \Bbb R\,:\, f'(x)<a\}$ is not equal to $\{x\in\Bbb R\,:\, f(x)<ax+b\}$ (whatever that $b$ is). Commented Oct 24, 2022 at 13:53
• Keep in mind that claiming, as you eventually do, that for all $f'$ and $a$, $f'^{-1}(-\infty,a)$ is open means claiming that $f'^{-1}(s,\infty)=(-f)'^{-1}(-\infty,-s)$ is open, and therefore that $f'^{-1}(s,a)=(f'^{-1}(s,\infty))\cap(f'^{-1}(-\infty,a))$ is open, and therefore that $f'$ is continuous, which is false. Commented Oct 24, 2022 at 13:58
• @Sven-OleBehrend that's right, can we modify it this way so that we can get a definite integral : $\int_{\alpha}^{t} f'(x) dx < \int_{\alpha}^{t} a dx$ implies $f(t) - f(\alpha) < at - \alpha$ . Let : $c = -\alpha + f(\alpha)$ then : $f(t) < at + c$ . More precisely, we can always pick an alpha for every $a$ such that the inequality is correct. Because the inequality property is true for definite integrals. Commented Oct 24, 2022 at 14:21
• @SassatelliGiulio I'm only claiming that the preimage of (−∞,a) by f' is Borel, not necessarily open or closed. Closed and clopen sets are Borel too. Commented Oct 24, 2022 at 14:22