Is the Set of Differentiable Points of a Function a Borel Set?

[Fermat1]

Let $f:R->R$ be a continuous function. Prove that set of points $f$ is differentiable at is a borel set.

I need to get to this set by union/intersection of intervals but how? I guess I'm missing a theorem about differentiable points and cotinuity

Thanks

 

[Euge]

Let $f:R->R$ be a continuous function. Prove that set of points $f$ is differentiable at is a borel set.

I need to get to this set by union/intersection of intervals but how? I guess I'm missing a theorem about differentiable points and cotinuity

Thanks

Let $X$ denote the set of points $x$ in $\Bbb R$ such that $f$ is differentiable at $x$. Then $x \in X$ if and only if the limit $\lim_{n\to \infty} f_n(x)$ exists, where

$$f_n(x) = n(f(x + 1/n) - f(x)).$$

Since $f$ is continuous, $f_n$ is a sequence of continuous (hence measurable) functions. So the set

$$X = \{x \in \Bbb R: \lim_{n\to \infty} f_n(x)\; \text{exists}\}$$

is Borel.

 

[Opalg]

Let $X$ denote the set of points $x$ in $\Bbb R$ such that $f$ is differentiable at $x$. Then $x \in X$ if and only if the limit $\lim_{n\to \infty} f_n(x)$ exists, where

$$f_n(x) = n(f(x + 1/n) - f(x)).$$

Since $f$ is continuous, $f_n$ is a sequence of continuous (hence measurable) functions. So the set

$$X = \{x \in \Bbb R: \lim_{n\to \infty} f_n(x)\; \text{exists}\}$$

is Borel.

That argument ought to work. But I am uneasy about it.

Example: Let $f(x) = x\sin\bigl(\frac{\pi}x \bigr)$, with $f(0) = 0.$ Then $f$ is continuous, but not differentiable at $0$. Yet $n(f(0 + 1/n) - f(0)) = n(\sin(n\pi) - \sin0) = 0.$ So $$\lim_{n\to \infty} f_n(0)$$ exists, despite $f$ not being differentiable at $0$.

 

[Euge]

That argument ought to work. But I am uneasy about it.

Example: Let $f(x) = x\sin\bigl(\frac{\pi}x \bigr)$, with $f(0) = 0.$ Then $f$ is continuous, but not differentiable at $0$. Yet $n(f(0 + 1/n) - f(0)) = n(\sin(n\pi) - \sin0) = 0.$ So $$\lim_{n\to \infty} f_n(0)$$ exists, despite $f$ not being differentiable at $0$.

Yes, that's right. What I gave is too simplified. Rather, $X$ can be equated with the set $\cap_{m \ge 1} \cup_{n \ge 1} X(m,n)$, where

$$ X(m,n) = \cap_{0 < |h| < \frac1{n}} \cap_{0 < |k| < \frac1{n}} \{x\in \Bbb R: |\Delta_hf(x) - \Delta_kf(x)| \le \frac1{m}\} $$

and $\Delta_uf(x)$ denotes the difference quotient $(f(x + u) - f(x))/u$. Since $f$ is continuous, $\Delta_uf$ is continuous for every $u$, so the sets

$$ \{x\in \Bbb R: |\Delta_hf(x) - \Delta_kf(x)| \le \frac1{m}\}$$

are all closed. Then the sets $X(m,n)$ are all closed. Consequently, for every $m$, the union $\cup_{n \ge 1} X(m,n)$ is Borel. Therefore, $X$ is the countable intersection of Borel sets, which shows that $X$ is Borel.