The integral function $F(x) := \int_a^x f$ of an integrable $f$

Let $a < b$ be reals and let $f : [a,b] \to \mathbb{R}$ be integrable (The lower and upper Darboux integrals of a bounded $f$ on $[a,b]$ as $\sup_P L(f,P)$ and $\inf_P U(f,P)$, Darboux integrability as their equality, and the notation $\int_a^b f$). The integral function of $f$ with base point $a$ is

$F : [a,b] \to \mathbb{R}, \qquad F(x) \;:=\; \int_a^x f .$

It is a genuine function, and that has to be checked. For $x \in (a,b]$ the restriction of $f$ to $[a,x]$ is integrable, by A function integrable on $[a,b]$ is integrable on every closed subinterval applied with $c := a$ and $d := x$, so $\int_a^x f$ names a single real number (The lower and upper Darboux integrals of a bounded $f$ on $[a,b]$ as $\sup_P L(f,P)$ and $\inf_P U(f,P)$, Darboux integrability as their equality, and the notation $\int_a^b f$). For $x = a$ the symbol $\int_a^a f$ is $0$ by The integral with oriented limits: $\int_a^a f := 0$ and $\int_b^a f := -\int_a^b f$. So $F(x)$ is defined at every point of $[a,b]$ and

$F(a) \;=\; 0 .$

More generally, for any base point $c \in [a,b]$ the function $x \mapsto \int_c^x f$ is defined on the whole of $[a,b]$, the integral being the oriented one of The integral with oriented limits: $\int_a^a f := 0$ and $\int_b^a f := -\int_a^b f$ when $x < c$; the case $c = a$ is written $F$ above and is the one used unless another base point is named.

The two identities used throughout

Increments are integrals. For all $x, y \in [a,b]$, in either order,

$F(y) - F(x) \;=\; \int_x^y f .$

This is claim 3 of For $a<c<b$: $f$ is integrable on $[a,b]$ if and only if it is integrable on $[a,c]$ and on $[c,b]$, and then $\int_a^b f = \int_a^c f + \int_c^b f$; with the oriented form for arbitrary $a,b,c$ applied to the three points $a$, $x$, $y$: it gives $\int_a^x f + \int_x^y f = \int_a^y f$, that is $F(x) + \int_x^y f = F(y)$. No ordering of $x$ and $y$ is assumed, and the degenerate cases $x = y$, $x = a$ and $y = a$ are included, since claim 3 is stated for arbitrary points.

Changing the base point changes $F$ by a constant. If $c \in [a,b]$ and $F_c(x) := \int_c^x f$, then for every $x \in [a,b]$

$F_c(x) \;=\; F(x) - F(c) ,$

again by claim 3 of For $a<c<b$: $f$ is integrable on $[a,b]$ if and only if it is integrable on $[a,c]$ and on $[c,b]$, and then $\int_a^b f = \int_a^c f + \int_c^b f$; with the oriented form for arbitrary $a,b,c$ at the points $a$, $c$, $x$. So the family of integral functions of $f$ is one function up to an additive constant.

• $F$ exists for every integrable $f$, whether or not $f$ has a primitive. Nothing in the definition asks $f$ to be continuous anywhere, and nothing here claims $F' = f$. The two statements about $F'$ that this page does prove are: $F$ is always Lipschitz (The integral function of a bounded integrable $f$ is Lipschitz, hence uniformly continuous), and $F'(c) = f(c)$ at every point $c$ where $f$ is continuous (The first fundamental theorem: if $f$ is integrable on $[a,b]$ and continuous at $c$, then $F'(c) = f(c)$; in particular a continuous $f$ has $F$ as a primitive).

• $F$ need not be a primitive of $f$. At a discontinuity of $f$ the derivative $F'$ may fail to exist, or may exist and differ from $f$; both possibilities are exhibited on the companion page, by The sign function is Riemann integrable on $[-1,1]$ and has no primitive there ↗ and FALSE: for every integrable $f$ on $[a,b]$, the integral function $F(x)=\int_a^x f$ satisfies $F' = f$ on $[a,b]$ ↗. That is the honest content of the phrase "the integral function", and it is why it is not called "the primitive" here.

• Why the base point is part of the data and the notation suppresses it. The symbol $F$ hides its dependence on $a$, as is customary; the identity $F_c = F - F(c)$ above is what makes the suppression harmless, since every statement below about $F$ is about its increments, which do not see the base point at all.