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

$F:[a,b]\to \mathbb{R},F(x):={\int }_a^{x}f\mathrm{.}$

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,

$F(y)-F(x)={\int }_x^{y}f\mathrm{.}$

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 $F^{\mathrm{\prime }}=f$. The two statements about $F^{\mathrm{\prime }}$ that this page does prove are: $F$ is always Lipschitz (The integral function of a bounded integrable $f$ is Lipschitz, hence uniformly continuous), and $F^{\mathrm{\prime }}(c)=f(c)$ at every point $c$ where $f$ is continuous (The first fundamental theorem: if $f$ is integrable on $[a,b]$ and continuous at $c$, then $F^{\mathrm{\prime }}(c)=f(c)$; 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 $F^{\mathrm{\prime }}$ 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 $F^{\mathrm{\prime }}=f$ 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.