FTC - is the differentiability of the antiderivative a necessary hypothesis?

I'm working on a proof for real analysis, and realized I'm not sure exactly when I can apply the fundamental theorem of calculus. Here is my book's statement of FTC, part 2:

If $f$ is differentiable on $[a,b]$ and $f'$ is integrable on $[a,b]$, then $$\int_{a}^{x}f'(t)dt= f(x)-f(a)$$

My question is... can one apply FTC to an arbitrary integrable function $g$ without knowing anything about $g$'s antiderivative? Is it safe to say that if a function is integrable, then its antiderivative is differentiable? I don't quite understand why $f'$'s differentiability is part of the hypothesis.

Can anyone explain or provide a counter-example (i.e. a function that is integrable but for which FTC cannot be used)?

Many thanks!

• I think you're referring to the first part of FTC - that is consider $G(x) = \int_{a}^{x} g(x) \, \mathrm{d} x$ then we have $G$ is differentiable and $G'(x) = g(x)$. – DanZimm Mar 4 '15 at 7:08

Counterexample of "if a function is integrable, then its antiderivative is differentiable": $$f(x)=\cases{0 & x\in[0,1]\cr 1 & x\in(1,2]}$$ is integrable but $$F(x)=\int_0^x f$$ isn't differentiable at $x=1$.

This is a common enough problem for students of analysis. I think an elaboration might help some patient reader out there, so I offer this.

Let's clarify the language a bit.

Definition 1. A function $F$ is an antiderivative of a function $f$ on an interval $I$ if $F'(x)=f(x)$ for every $x$ in the interval $I$. [NOTE: every!]

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Definition 2. A function $F$ is an indefinite integral [in a specified sense] of a function $f$ on an interval $[a,b]$ if $f$ is integrable on $[a,b]$ [in that sense] and $$\int_c^d f(x)\,dx = F(d)-F(c)$$ for all $a\leq c<d\leq b$.

Some clarifications. In the first definition, with a closed interval $[a,b]$, we would probably specify one-sided derivatives at the end points [i.e., $F_+'(a)=f(a)$ and $F_-'(b)=f(b$)]. Also if $F$ is an antiderivative then so is $F+C$ for any constant $C$ and if $F$ is an indefinite integral [some sense] then so is $F+C$ for any constant $C$.

Also we could define an indefinite integral instead as any function of the form $$F(x)=\int_a^x f(t)\,dx + C$$ for any constant $C$ and function $f$ integrable on $[a,b]$.

So much for definitions. What is the connection?

Newton integral. For all of the 18th century (and beyond) integration was just "antidifferentiation" and there is hardly any difference between an antiderivative and an indefinite integral.

Define a function to be Newton integrable on $[a,b]$ if there is an antiderivative $F$ for $f$ on $[a,b]$ and set the value $\int_a^bf(x)\,dx=F(b)-F(a)$.

So the two ideas are identical: an antiderivative is exactly an indefinite integral [Newton sense].

Riemann integral (we are up to the mid 19th century now).

(a) Continuous functions. Every function $f$ that is continuous on an interval $[a,b]$ is Riemann integrable there and its indefinite integral [Riemann sense] is an antiderivative.

So again the two ideas are identical: an antiderivative of a continuous function is exactly an indefinite integral [Riemann sense].

(b) Discontinuous functions. Here is where the two ideas part company (as one of the other answers illustrates).

There are discontinuous functions on an interval $[a,b]$ that have an antiderivative, but are not Riemann integrable. (Unbounded functions provide examples, but there are also bounded examples.)

There are discontinuous functions on an interval $[a,b]$ that are Riemann integrable but do not have an antiderivative. If $F$ is an indefinite Riemann integral, i.e., if $$F(x)=\int_a^x f(t)\,dx,$$ then $F'(x)$ must exist at most points and $F'(x)=f(x)$ at most points, but an antiderivative has to be differentiable everywhere. If $f$ has a discontinuity at a point $x_0$ it is possible that $F'(x_0)$ does not exist and it is possible that it does exist but that $F'(x_0)\not=f(x_0)$.

Other integrals. One needs to go past the Newton and Riemann integrals for many purposes of analysis. Thus there is the improper Riemann integral, the Lebesgue integral and the Denjoy-Perron integral (among many others).

What is the connection between antiderivatives and indefinite integrals in these more general senses?

Since these all include the Riemann integral it is clear that indefinite integrals do not have to be antiderivatives ($F'(x)=f(x)$ can fail at many points [but not all points]).

Are antiderivatives also always indefinite integrals in any sense? No for improper Riemann and Lebesgue integrals. It is possible to find a differentiable function $F$ on an interval $[a,b]$ so that $F'$ is not integrable in either sense.

The more mysterious Denjoy-Perron integral was designed explicity to handle this problem. If $F$ is differentiable on interval $[a,b]$ then even though $F'$ may fail to be Riemann integrable or Lebesgue integrable, it will be Denjoy-Perron integrable.

Summary: Is an antiderivative $F$ an indefinite integral [in some sense]? Maybe, but you have to check whether $F'$ is integrable in that sense (e.g., if $F'$ is continuous, then $F'$ is Riemann integrable). For the Riemann and Lebesgue integrals --maybe it is integrable, maybe not. The Denjoy-Perron integral is the only one here that guarantees integrability.

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Summary: Is an indefinite integral $F$ [in some sense] also an antiderivative? Probably not. You will find that $F'(x)=f(x)$ at most points, but likely not at every point (unless $f$ is continuous).

In general, the Riemann integral of a map, like the one provided in the counterexample above, is continous. The integral of a continuous map is continuously differentiable.

In general, the value of the FTC is to provide a relationship between the derivative and the integral of a function, bridging integral and derivative calculus.