Theorem: Continuous Real Function Differentiable on Borel Set

Let $\map \BB {\R, \size {\, \cdot \,} }$ be the Borel Sigma-Algebra on $\R$ with the usual topology.
Let $f: \R \to \R$ be a continuous real function.
Let $\map D f$ be the set of all points at which $f$ is differentiable.
Then $\map D f$ is a Borel Set with respect to $\map \BB {\R, \size {\, \cdot \,} }$.

Proof:
By the definition of derivative:
:$\map {f'} x$ exists 
{{iff}}:
:$\ds \lim_{h \mathop \to 0} \frac {\map f {x + h} - \map f x} h = L$ 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, $\ds \sum_{n \mathop \ge 0} \paren {y_{n + 1} - y_n}$ converges.
The difference of successive terms makes this a Telescoping Series:
:$\ds \sum_{n \mathop \ge 0} \paren {y_{n + 1} - y_n} = -y_0 +  \lim_{n \mathop \to \infty} y_n$
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:
:$\ds \bigcap_n \bigcup_{\delta_n} \bigcup_{y_n} \bigcap_h \set {x \in \R: \size {\dfrac {\map f {x + h} - \map f x} h - y_n} < 2^{-n} }$
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:Borel Sets
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