User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)

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Theorem
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




 * $\displaystyle \lim_{h \mathop \to 0} \frac{\map f {x+h}-\map f x }{h} - L$ exists

where $L = \map{f'}{x}$.

For ease of presentation, denote:


 * $\map{f}{x+h}-\map{f}{x} := \Delta f$

By the definition of limit, this statement is:


 * $\displaystyle \exists L \in \R: \forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall h \in \R \setminus \set{0}: \size{h}<\delta \implies \size{ \dfrac{\Delta f}{h} -L} < \epsilon$

From Limit with Rational $\epsilon$ and $\delta$ and Limit with $\epsilon$ Powers of $2$, we can instead consider:


 * $\displaystyle \exists L \in \R: \forall n \in \N: \exists \delta \in \Q_{>0}: \forall h \in \Q \setminus \set{0}: \size{h}<\delta \implies \size{\frac{\Delta f}{h} -L} < 2^{-n}$.

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:

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. Call it $y$.

Then:

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:


 * $\displaystyle \size{\Delta f - hy } \le \size{h} \paren{2^{-n} + \size{y_n -y} }$

Equivalently:


 * $\displaystyle \size{\frac{\map{f}{x+h}-\map{f}{x}}{h} - y } \le \paren{2^{-n} + \size{y_n -y} }$

The right hand side tends to $0$ as $n \to \infty$, as $2^{-n} \to 0$ and $y_n \to y$.

Thus the left-hand side 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 $y = f'(x)$ by the definition of derivative, and we have shown that our assumptions imply the existence of $f'(x)$.

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:


 * $\displaystyle \bigcap_{n} \bigcup_{\delta_n} \bigcup_{y_n} \bigcap_h \set{x \in \R: \size{\dfrac{\map{f}{x+h}-\map{f}{x}}{h} -y_n}<2^{-n}}$

... and each set so indexed is open, because $f$ is continuous by hypothesis.