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Such a function does not exist.

Indeed, let $f\colon[a,b]\to\mathbb R$ be an increasing differentiable function. Then $$f(x)-f(a)=(HK)\int_a^x f'(t)\,dt$$ for all $x$, where $(HK)\int$ is the Henstock–Kurzweil integral. In particular, $f$ is Henstock–Kurzweil integrable on $[a,b]$. Therefore and because $f'\ge0$, $f'$ is Lebesgue integrable on $[a,b]$, and hence $$f(x)-f(a)=(L)\int_a^x f'(t)\,dt$$ for all $x$, where $(L)\int$ is the Lebesgue integral.

So, $f$ must be absolutely continuous.

Such a function does not exist.

Indeed, let $f\colon[a,b]\to\mathbb R$ be an increasing differentiable function. Then $$f(x)-f(a)=(HK)\int_a^x f'(t)\,dt$$ for all $x$, where $(HK)\int$ is the Henstock–Kurzweil integral. In particular, $f'$ is Henstock–Kurzweil integrable on $[a,b]$. Therefore and because $f'\ge0$, $f'$ is Lebesgue integrable on $[a,b]$, and hence $$f(x)-f(a)=(L)\int_a^x f'(t)\,dt$$ for all $x$, where $(L)\int$ is the Lebesgue integral.

So, $f$ must be absolutely continuous.

Such a function does not exist.

Indeed, let $f\colon[a,b]\to\mathbb R$ be an increasing differentiable function. Then $$f(x)-f(a)=(HK)\int_a^x f'(t)\,dt$$ for all $x$, where $(HK)\int$ is the Henstock–Kurzweil integral. In particular, $f$ is Henstock–Kurzweil integrable on $[a,b]$. Therefore and because $f'\ge0$, $f'$ is Lebesgue integrable on $[a,b]$, and hence $$f(x)-f(a)=(L)\int_a^x f'(t)\,dt$$ for all $x$, where $(L)\int$ is the Lebesgue integral.

So, $f$ is absolutely continuous.

Such a function does not exist.

Indeed, let $f\colon[a,b]\to\mathbb R$ be an increasing differentiable function. Then $$f(x)-f(a)=(HK)\int_a^x f'(t)\,dt$$ for all $x$, where $(HK)\int$ is the Henstock–Kurzweil integral. In particular, $f$ is Henstock–Kurzweil integrable on $[a,b]$. Therefore and because $f'\ge0$, $f'$ is Lebesgue integrable on $[a,b]$, and hence $$f(x)-f(a)=(L)\int_a^x f'(t)\,dt$$ for all $x$, where $(L)\int$ is the Lebesgue integral.

So, $f$ must be absolutely continuous.

Let $f\colon[a,b]\to\mathbb R$ be an increasing differentiable function. Then $$f(x)-f(a)=(HK)\int_a^x f'(t)\,dt$$ for all $x$, where $(HK)\int$ is the Henstock–Kurzweil integral. Therefore and because $f'\ge0$, $f'$ is Lebesgue integrable and hence $$f(x)-f(a)=(L)\int_a^x f'(t)\,dt$$ for all $x$, where $(L)\int$ is the Lebesgue integral.

So, $f$ is absolutely continuous.

Such a function does not exist.

Indeed, let $f\colon[a,b]\to\mathbb R$ be an increasing differentiable function. Then $$f(x)-f(a)=(HK)\int_a^x f'(t)\,dt$$ for all $x$, where $(HK)\int$ is the Henstock–Kurzweil integral. In particular, $f$ is Henstock–Kurzweil integrable on $[a,b]$. Therefore and because $f'\ge0$, $f'$ is Lebesgue integrable on $[a,b]$, and hence $$f(x)-f(a)=(L)\int_a^x f'(t)\,dt$$ for all $x$, where $(L)\int$ is the Lebesgue integral.

So, $f$ is absolutely continuous.