Why does $f$ in $\int f(x)\, dx$ have to be continuous?

268 Views Asked by At

Let $f:(a\,..b) \rightarrow \mathbb R$ and $F:[a\,..b]\rightarrow \mathbb R$ be two real functions. Let $F$ be differentiable on $(a\,..b)$ and continuous on $[a\,..b]$. Let $f$ be continuous on $(a\,..b)$. Then $F$ is called an antiderivative of $f$ with respect to $x$ iff $\forall x \in(a\,..b):F'(x)=f(x)$.

Why, in this definition, does $f$ need to be continuous on $(a\,..b)$? Why isn't $F$ considered an antiderivative of $f$ when $\forall x \in (a\,..b):F'(x)=f(x)$ but $f$ isn't continuous?

1

There are 1 best solutions below

0
On BEST ANSWER

I suppose you can call $F$ an antiderivative of $f$ even when $f$ isn't continuous. That's just a matter of terminology. I'm not sure what the conventions are.

However, note that if $f$ is continuous, it is very easy to prove that the antiderivative $F$ actually exists. You just take $F(x) = \int_a^x f$; the fact that $F'(x) = f(x)$ follows by observing that $(F(x+h)-F(x))/h = (1/h)\int_x^{x+h} f$ which tends to $f(x)$ as $h \to 0$ by continuity of $f$.

Also, note that for $f$ Lebesgue integrable but not necessarily continuous on $[a,b]$, the function $F(x) = \int_a^x f$ obeys $F'(x) = f(x)$ almost everywhere. So the notion of antiderivatives does apply to discontinuous $f$. See Theorem 7.11 in Rudin's Real and Complex Analysis.