Derivative of Differentiable Function on Open Interval is Baire Function

Theorem
Let $I = \openint a b$ be a open interval.

Let $f : I \to \R$ be a differentiable function.

Let $f' : I \to \R$ be the derivative of $f$.

Then $f'$ is a Baire function.

Proof
For each $n \in \N$, let $f_n : I \to \R$ be a function with:
 * $\ds \map {f_n} x = n \paren {\map f {\min \set {x + \frac 1 n, \frac {b + x} 2} } - \map f x}$

for each $x$.

Note that from:
 * Minimum of Finitely Many Continuous Real Functions is Continuous
 * Composite of Continuous Mappings is Continuous
 * Combined Sum Rule for Continuous Real Functions

we have:
 * $f_n$ is continuous for each $n$.

We aim to show that $f_n$ converges pointwise to $f'$.

Fix $x \in I$.

Let $\epsilon$ be a positive real number.

Since $f$ is differentiable with derivative $f'$, there exists $\delta > 0$ such that:
 * $\ds \size {\frac {\map f {x + h} - \map f x} h - \map {f'} x} < \epsilon$

for all real $h$ with $0 < \size h < \delta$.

Let $N$ be the least natural number such that:
 * $\ds x + \frac 1 N < \frac {b + x} 2$

That is:
 * $\ds N > \frac 2 {b - x} > 0$

Then, for $n > N$ we have:
 * $\ds \min \set {x + \frac 1 n, \frac {b + x} 2} = x + \frac 1 n$

Then, for $n > \max \set {1/\delta, N}$ we also have $0 < 1/n < \delta$, and so:

By the definition of pointwise convergence:
 * $f_n$ converges pointwise to $f'$.

Since $\sequence {f_n}$ is a sequence of continuous real functions converging pointwise to $f'$, we have:
 * $f'$ is a Baire function.