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

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

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

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

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

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

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

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

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 \(\displaystyle \lim_{n\to \infty} f_n(0)\) 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 \(\displaystyle \lim_{n\to \infty} f_n(0)\) 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

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

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

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

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.

 

[blue_raver22]

for your question. To prove that the set of points where $f$ is differentiable is a Borel set, we can use the fact that continuous functions are Borel measurable. This means that the preimage of any Borel set under $f$ is also a Borel set.

Let $A$ be the set of points where $f$ is differentiable. We can write $A$ as the union of two sets: $A_1$, the set of points where $f$ is differentiable from the left, and $A_2$, the set of points where $f$ is differentiable from the right.

To show that $A_1$ is a Borel set, we can use the fact that the set of points where a function is differentiable from the left is the same as the set of points where the left-hand derivative exists. Since $f$ is continuous, the left-hand derivative exists at every point and is equal to the limit of the function from the left. This means that $A_1$ can be written as the preimage of the set of real numbers, which is a Borel set.

Similarly, $A_2$ can be written as the preimage of the set of real numbers, which is also a Borel set.

Since $A$ is the union of two Borel sets, it is also a Borel set. Therefore, the set of points where $f$ is differentiable is a Borel set.