# For an a.e. differentiable continuous function $f$ with $f'$ integrable, is it true in general that $f(b)-f(a) \geq \int_a^b f'(x) dx$?

I am aware that the fundamental theorem of calculus fails for functions which are a.e. differentiable but not absolutely continuous, such as the Cantor function.

However, I wonder if a general inequality still holds in one direction. That is, for any continuous $$f : [a,b] \to \mathbb{R}$$ that is differentiable almost everywhere on the interval $$[a,b]$$ with $$f'$$ being integrable, I wonder if we have in general $$$$f(b)-f(a) \geq \int_a^b f'(x)dx.$$$$

• If the inequality holds for $f$ and $-f$ you get equality. So if you do not expect equality you should assume that $f'$ has constant sign. Sep 23, 2023 at 10:57

The Cantor function allows you to go up or down without being accounted by the derivative.

Let $$g$$ be the Cantor function and define $$f(x)=\begin{cases} x,&\ 0\leq x \leq \frac12\\[0.2cm] 1-g(x),&\ \frac12 Then $$f$$ is continuous, it is differentiable almost everywhere, $$f(1)=f(0)$$, and $$\int_0^1f'=\int_0^{1/2}1=\frac12.$$

• Perhaps the inequality holds if $f$ is increasing. Sep 23, 2023 at 11:13

If $$f$$ is decreasing, this doesn’t hold, but it’s true if $$f$$ is increasing. Indeed, the difference quotients $$f_h(x) = \frac1h(f(x+h)-f(x))$$ converge a.e. to $$f’(x)$$ and are nonnegative, so by Fatou’s lemma,

$$\int_a^b f’(x)dx \leq \liminf_{h\to 0} \int_a^b f_h(x) dx$$

But

\begin{align} \int_a^b f_h(x) dx &= \frac1h \int_{a+h}^{b+h} f(x)dx - \frac1h \int_a^b f(x)dx \\ &= \frac1h \int_{b}^{b+h} f(x)dx - \frac1h \int_a^{a+h} f(x)dx\end{align}

Since $$f$$ is continuous, the above converges to $$f(b)-f(a)$$.