By the definition of derivative: :$\map {f'} x$ exists  {{iff}}: :$\displaystyle \lim_{h \mathop \to 0} \frac {\map f {x + h} - \map f x} h = L$ exists where $L = \map {f'} x$. For ease of presentation, denote: :$\Delta f := \map f {x + h} - \map f x$ Since $f$ is continuous by hypothesis, so too is $\dfrac {\Delta f} h$ for $h \ne 0$. Thus we will prove an inequality for $\dfrac {\Delta f} h$ for all rational $h$. The inequality will then extend to all real $h$ by Rationals are Everywhere Dense in Topological Space of Reals. By the definition of limit, the claim that the derivative exists is: :$\exists L \in \R: \forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall h \in \R \setminus \set 0: \size h < \delta \implies \size {\dfrac {\Delta f} h - L} < \epsilon$ From Limit with Rational $\epsilon$ and $\delta$ and Limit with $\epsilon$ Powers of $2$, we can instead consider: :$\exists L \in \R: \forall n \in \N: \exists \delta \in \Q_{>0}: \forall h \in \Q \setminus \set 0: \size h < \delta \implies \size {\dfrac {\Delta f} h - L} < 2^{-n}$ Let $\set {\delta_n}$ be an enumeration of all such $\delta$s considered. Assume the following: :$\forall n \in \N: \exists \delta_n \in \Q_{>0}: \exists y_n \in \Q: \forall h \in \closedint {-\delta_n} {\delta_n}: \size {\map f {x + h} - \map f x - h y_n} \le \size h 2^{-n}$ for some rational sequence $\sequence {y_n}_{n \mathop \in \N}$. Then:  {{begin-eqn}} {{eqn | l = \size {y_{n + 1} - y_n}       | r = \size {y_{n + 1} + \frac {\Delta f} h - \frac {\Delta f} h - y_n} }} {{eqn | o = \le       | r = \size {\frac {\Delta f} h - y_{n + 1} } + \size {\frac {\Delta f} h - y_n}       | c = Triangle Inequality for Real Numbers }} {{eqn | o = \le       | r = 2^{-\paren {n + 1} } + 2^{-n}       | c = By Hypothesis }} {{eqn | o = <       | r = 2^{-n + 1} }} {{end-eqn}} By the Comparison Test, $\displaystyle \sum_{n \mathop \ge 0} \paren {y_{n + 1} - y_n}$ converges. The difference of successive terms makes this a Telescoping Series: :$\displaystyle \sum_{n \mathop \ge 0} \paren {y_{n + 1} - y_n} = -y_0 +  \lim_{n \mathop \to \infty} y_n$ so $y_n$ converges to a limit, which will be called $y$. Then: {{begin-eqn}} {{eqn | l = \size {\Delta f - h y}       | r = \size {\Delta f - h y_n + h y_n - h y} }} {{eqn | o = \le       | r = \size {\Delta f - h y_n} + \size {h y_n - h y}       | c = Triangle Inequality for Real Numbers }} {{eqn | o = \le        | r = \size {h 2^{-n} } + \size {h y_n - h y} }} {{eqn | r = \size h \paren {2^{-n} + \size {y_n - y} } }} {{end-eqn}} We have thus shown that the assumption that: :$\forall n \in \N: \exists \delta_n \in \Q_{>0}: \exists y_n \in \Q: \forall h \in \closedint {-\delta_n} {\delta_n}: \size{ \map f {x + h} -\map f x - h y_n } \le \size h 2^{-n}$ ... implies that for all $n \in \N, -\delta_n < h < \delta_n$, there is a real number $y$ such that: :$\size {\Delta f - h y} \le \size h \paren {2^{-n} + \size {y_n - y} }$ Equivalently: :$\size {\dfrac {\map f {x + h} - \map f x} h - y} \le \paren {2^{-n} + \size {y_n -y} }$ The {{RHS}} tends to $0$ as $n \to \infty$, as $2^{-n} \to 0$ and $y_n \to y$. Thus the {{LHS}} also tends to $0$ as $n \to \infty$. Thus $\dfrac{\map f {x + h} - \map f x} h \to y$ as $n \to \infty$. Then $y = \map {f'} x$ by the definition of derivative, and we have shown that our assumptions imply the existence of $\map {f'} x$. The set of all $x \in \R$ satisfying: :$\forall n \in \N: \exists \delta_n \in \Q_{>0}: \exists y_n \in \Q: \forall h \in \closedint {-\delta_n} {\delta_n}: \size {\map f {x + h} - \map f x - h y_n} \le \size h 2^{-n}$ is: :$\displaystyle \bigcap_n \bigcup_{\delta_n} \bigcup_{y_n} \bigcap_h \set {x \in \R: \size {\dfrac {\map f {x + h} - \map f x} h - y_n} < 2^{-n} }$ By hypothesis, $f$ is continuous. Hence each set so indexed is open.  So it is seen that $\map D f$ has been constructed from countably many intersections and unions of open sets. Hence $\map D f$ is a Borel Set by definition. {{qed}} {{AoC|Countable Union of Countable Sets is Countable}} Category:Continuity Category:Sigma-Algebras nvssjxol5pzkyp3t4imn8uj9pdf0g29