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.

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:

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:

$\ds \size {\map {f_n} x - \map {f'} x}$

$=$

$\ds \size {n \paren {\map f {x + \frac 1 n} - \map f x} - \map {f'} x}$

$\ds $

$=$

$\ds \size {\frac {\map f {x + \frac 1 n} - \map f x} {\frac 1 n} - \map {f'} x}$

$\ds $

$<$

$\ds \epsilon$

By the definition of pointwise convergence:

$\blacksquare$