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 $f'(x)<a\}$ is Borel.

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

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

$f'^{-1}(- ∞ ,a)= \{x\in \mathbb{R}$ s.t $f(x)-ax<b\} = g^{-1}(- ∞ ,b)$ 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

  • 1
    $\begingroup$ 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 $F \not < G$. $\endgroup$ Commented Oct 24, 2022 at 13:51
  • $\begingroup$ $\{x\in \Bbb R\,:\, f'(x)<a\}$ is not equal to $\{x\in\Bbb R\,:\, f(x)<ax+b\}$ (whatever that $b$ is). $\endgroup$ Commented Oct 24, 2022 at 13:53
  • $\begingroup$ 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. $\endgroup$ Commented Oct 24, 2022 at 13:58
  • $\begingroup$ @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. $\endgroup$ Commented Oct 24, 2022 at 14:21
  • $\begingroup$ @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. $\endgroup$ Commented Oct 24, 2022 at 14:22


You must log in to answer this question.

Browse other questions tagged .