Integrable function of which the antiderivative is not equal to the integral

Question

Is there a Riemann-integrable function $f: [a,b]\to\mathbb{R}$ such that $f$ has an antiderivative which is not equal to $t \mapsto \int_a^t f(x)dx +c$ for any $c\in\mathbb{R}$?

Answer 1

The answer is no, but at the moment I need some measure theory to prove it.

Because $f$ is Riemann integrable on $[a,b],$ $f$ is continuous a.e. there. Therefore, by the FTC, $\int_a^x f(t)\,dt$ has derivative equal to $f(x)$ for a.e. $x.$

Let $g$ be an antiderivative for $f$ on $[a,b].$ Define

$$h(x) = g(x) - \int_a^x f(t)\,dt.$$

Then $h'(x) = f(x)-f(x)=0$ for a.e. $x.$ But note that $g,$ having bounded derivative, is Lipschitz. So is $\int_a^x f(t)\,dt.$ Thus $h$ is Lipschitz, hence absolutely continuous. An absolutely continuous function whose derivative equals $0$ a.e. is constant. Hence $h$ is constant as desired.

Answer 2

The assumption is that $f:[a,b] \to \mathbb{R}$ has an antiderivative. This means there is a differentiable function $F:[a,b] \to \mathbb{R}$ such that $F'(x) = f(x)$ for all $x \in [a,b]$. Obviously $F$ is continuous on $[a,b]$ since it is differentiable, and $F(a)$ is well defined. However, we make no assumption that $f$ is continuous at every point in $[a,b]$, only that it is Riemann integrable.

With $t \in (a,b]$ fixed and a partition $P: a = x_0 < x_1 < \ldots < x_n = t$, it follows from the MVT that there exist intermediate points $\xi_j \in (x_{j-1},x_j)$ such that

$$F(t) - F(a) = \sum_{j=1}^n (\, F(x_j) - F(x_{j-1}) \, ) = \sum_{j=1}^n F'(\xi_j)(x_j - x_{j-1})= \sum_{j=1}^n f(\xi_j)(x- x_{j-1})$$

Thus,

$$\left|F(t) - F(a) - \int_a^tf(x) \, dx\right| = \left| \sum_{j=1}^n f(\xi_j)(x- x_{j-1}) - \int_a^tf(x) \, dx\right|$$

Of course, since $f$ is Riemann integrable, the partition $P$ can always be selected so that, regardless of the choice of intermediate points, the RHS is smaller than any $\epsilon > 0$. Since the LHS does not depend on which partition is selected, we have for any $\epsilon > 0$

$$\left|F(t) - F(a) - \int_a^tf(x) \, dx\right| < \epsilon$$

Since $\epsilon$ is arbitrary, we must have for all $t \in [a,b]$ and $c = F(a)$,

$$F(t) = c + \int_a^tf(x) \, dx$$