# Is Spivak wrong about this counterexample? $f$ integrable on $[-1,1]$, $F=\int_{-1}^xf$, $f$ differentiable at $0$, but $F'$ not continuous at $0$

Let $$f$$ be Riemann integrable on $$[a,b]$$, let $$c\in(a,b)$$, and let $$F(x)=\int_a^xf$$ for $$x\in[a,b]$$.

Exercise 3b in Chapter 14 of Spivak's Calculus (3rd edition) asks to give a proof or provide a counterexample of this statement:

If $$f$$ is differentiable at $$c$$, then $$F'$$ is continuous at $$c$$.

In his solutions manual, Spivak claims to give a counterexample. His function is $$f:[-1,1]\to\mathbb{R}$$ defined by

$$f(x)=\begin{cases}0&x=0\\1&x\in\left\{-1,1\right\}\\\frac{1}{n^2}&\frac{1}{n}\leq |x|<\frac{1}{n-1}, n\geq2\end{cases}$$

Observe that $$f$$ has jump discontinuities at points of the form $$1/n$$ for nonzero integers $$n$$ and is continuous everywhere else.

Now certainly $$f$$ is Riemann integrable on $$[-1,1]$$ (because its sets of discontinuities is countable, thus measure zero). Moreover $$f$$ is indeed differentiable at $$0$$.

But I don't think $$F'$$ is discontinuous at $$0$$. On the contrary, here is my proof that $$F'$$ is continuous at $$0$$.

First, since $$f$$ is differentiable at $$0$$, $$f$$ is continuous at $$0$$; it follows by the fundamental theorem that $$F'$$ exists at $$0$$ and that $$F'(0)=f(0)$$. We know $$f(0)=0$$, so $$F'(0)=0$$.

It remains to show that $$\lim_{x\to0}F'(x)=0$$. Now this is a little delicate because, as Spivak points out, $$F'$$ is not even defined at $$1/n$$ for $$n$$ a nonzero integer. But that doesn't prevent us from calculating the limit of $$F'$$. The limit simply must be taken through the domain of $$F'$$.

Indeed, given $$\epsilon>0$$, take $$N$$ such that $$\frac{1}{N^2}<\epsilon$$, and pick $$\delta=\frac{1}{N}$$.

Then if $$0<|x|<\delta$$ and $$x$$ is in the domain of $$F'$$, it follows that $$|F'(x)|\leq\frac{1}{(N+1)^2}<\frac{1}{N^2}<\epsilon$$.

Therefore it is indeed true that

$$\lim_{x\to0}F'(x)=F'(0)$$

so $$F'$$ is continuous at $$0$$.

What's going on here? Is this just a matter of one's definition of the limit of a function on a set that isn't an interval? Spivak seems to think "If $$\lim_{x\to c}g(x)$$ exists then $$g$$ is defined on a deleted interval around $$c$$."

• That is indeed what is going on. See this question which quotes the definition of “limit” that Spivak is using, which requires the function to be defined in a deleted interval around the limit point. Commented Mar 9, 2021 at 2:09
• While it is true that one may generally define limits on any accumulation point of the domain, this is often not done in Calculus, and Spivak is not doing it here. See some of the discussion that occurs inter alia here about the two possible ways to define a limit. Commented Mar 9, 2021 at 2:11
• @ArturoMagidin: thank you! Commented Mar 9, 2021 at 2:18
• Even though the question I've asked is a boring one, turning on a technicality, there's a real question lurking just behind: I wonder if the claim "If $f$ is differentiable at $c$, then $F'$ is continuous at $c$" is actually true if you work with the more general definition of a limit. Commented Mar 9, 2021 at 2:27
• I hope someone comes up with a proof that no such example exists. Commented Mar 9, 2021 at 4:18

The answer to your question in the comments is yes: If $$f$$ is differentiable at $$c,$$ then $$F'$$ is continuous at $$c$$ within the domain of $$F'.$$

Proof: Suppose $$y_n\to c^+$$ (the case $$y_n\to c^-$$ is similar). Assume $$F'(y_n)$$ exists for each $$n.$$ Because $$f'(c)$$ exists, we can write

$$\tag 1 f(x) = f(c) +f'(c)(x-c) + \epsilon(x)(x-c),$$

where $$\epsilon(x)\to 0$$ as $$x\to c.$$ Actually we don't need the $$\epsilon\to 0$$ property. If you divide both sides of $$(1)$$ by $$x-c,$$ you'll see $$\epsilon(x)$$ is bounded on $$[a,b]\setminus \{c\},$$ say by $$M$$ in absolute value.

Fix $$n.$$ Then for small $$h>0,$$

$$\tag 2 \frac{F(y_n+h)-F(y_n)}{h} = \frac{1}{h}\int_{y_n}^{y_n+h} f(x)\,dx.$$

By $$(1),$$ $$(2)$$ equals

$$\tag 3 f(c) + \frac{1}{h}\int_{y_n}^{y_n+h}(f'(c)+\epsilon(x))(x-c)\,dx.$$

Now the second expression in $$(3)$$ is bounded above in absolute value by

$$(|f'(c)|+M)\frac{1}{h}\int_{y_n}^{y_n+h}(x-c)\,dx = (|f'(c)|+M)(y_n-c+h).$$

Thus

$$\left |\frac{F(y_n+h)-F(y_n)}{h} - f(c)\right |\le (|f'(c)|+M)(y_n-c+h).$$

Taking the limit as $$h\to 0$$ then gives

$$|F'(y_n) - f(c)| \le (|f'(c)|+M)(y_n-c).$$

Now letting $$n\to \infty,$$ we see

$$F'(y_n) \to f(c) = F'(c),$$

and we're done.

• Terrific! Thanks! An interesting subtlety in a problem I'd never thought about before. Commented Mar 9, 2021 at 20:04
• Sorry, realizing I don’t follow one point. If we define $\epsilon(x)$ as $(f(x)-f(c))/(x-c)-f’(c)$ for $x\neq c$ and $0$ at $c$, it’s obvious why $\epsilon$ is continuous at $c$, but don’t we need $\epsilon$ to be continuous away from $c$ too? (When we use FTC near the end, we need $\epsilon$ to be continuous at $y_n\neq c$.) I don’t see why that follows (it would be true if $f$ were continuous at $y_n$, but it need not be). Commented Mar 10, 2021 at 0:27
• Good point. I think it can be fixed, I'll have to get back to you on this.
– zhw.
Commented Mar 10, 2021 at 2:07
• Let us also inform @symplectomorphic about the update (via this comment). They will be glad to see the update. Commented Mar 10, 2021 at 3:34
• @ParamanandSingh Will do.
– zhw.
Commented Mar 10, 2021 at 3:36