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Can anyone give me a counterexample to the following statement:

Suppose $F \colon [0,1] \to \mathbb{R}$ is continuous and differentiable almost everywhere, then $F(b)-F(a)=\int_a^b F'(t)\, \text{d}t$.

I guess let the discontinuity be $\mathbb{Q} \cap[0,1]$ may help.

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  • $\begingroup$ Even if $F$ is differentiable everywhere it is not necessarily true that $F'$ is Lebesgue integrable. You can,however, add hypotheses to make this true, e.g., that $F'$ is bounded, or that $F$ has bounded variation, but in general the Lebesgue integral does not have a FTC for continuous almost everywhere differentiable or even for everywhere differentiable functions. $\endgroup$ – B. S. Thomson Jan 1 '16 at 1:25
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    $\begingroup$ Does this mean "{continuous and differentiable} almost everywhere", or "continuous and {differentiable almost everywhere}"? Presumably the latter since otherwise this is a triviality. But I think it would be better to phrase it unambiguously, perhaps by saying "continuous everywhere and differentiable almost everywhere". ${}\qquad{}$ $\endgroup$ – Michael Hardy Jan 1 '16 at 2:08
  • $\begingroup$ @MichaelHardy The latter (as the Cantor function below illustrates). I made the observation just to stress that it is not the "almost everywhere differentiability" that is interfering with the FTC. Even "everywhere differentiability" is not enough for the Lebesgue integral. Along that line, perhaps you recall that every measurable function is the a.e. derivative of some continuous function, so the FTC fails quite remarkably in many ways if you don't make the right assumptions. $\endgroup$ – B. S. Thomson Jan 1 '16 at 3:28
  • $\begingroup$ @B.S.Thomson Yes, these concepts are similar but not the same. I guess if $F$ is differentiable everywhere, then the counterexample may more hard to find. $\endgroup$ – Math Boy Jan 1 '16 at 5:23
  • $\begingroup$ The ambiguity in the statement (as highlighted by Michael Hardy) might be removed if the hypothesis was instead "$F$ ... is continuous [a.e.] on $[a,b]$ and differentiable [a.e.] on $(a,b)$" with whichever of the bracketed "a.e."s you intended. $\endgroup$ – Eric Towers Jan 6 '16 at 5:28
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The Cantor function is a counter-example. It's continuous and has derivative $0$ almost everywhere, yet $F(1)-F(0)=1-0=1 \neq 0=\int_0^10 \, dt=\int_0^1F'(t) \, dt$.

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  • $\begingroup$ Thanks! I forgot this most important "counterexample" function. $\endgroup$ – Math Boy Jan 1 '16 at 5:18
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    $\begingroup$ A last unnecessary comment: the identity $$F(b)-F(a)=\int_a^b F'(t)\, {d}t$$ can fail for two reasons: (i) both sides exist and are unequal or (ii) $F'$ (which is a measurable function defined a.e.) might fail to be integrable. The Cantor example is of the former type. But there are also examples for which $F'$ is unbounded and not integrable by any method. The Cantor example is the default answer. But it is useful to keep in mind that the property of being the a.e. derivative of a continuous function imposes no properties on a function: every measurable function has this property. $\endgroup$ – B. S. Thomson Jan 2 '16 at 0:17
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I harass Calculus students with this example (which one can easily scale and translate to $[0,1]$):

Let $f(x) = \tan^{-1}(x) + \tan^{-1}(1/x)$. Compute $f(1)$, $f(-1)$, and $f'(x)$. How is that possible?

They find $f(1) = \pi/2$, $f(-1) = -\pi/2$, and $f'(x) = 0$ (almost) everywhere. Which should be a good enough hint.

Worth noting that this $f$ is continuous (constant!) a.e. and differentiable a.e. I like this $f$ because the explanation of constancy is simple if one remembers one's Trig: $f$ is the sum of two (directed) complementary angles. The direction just reverses as we pass through $0$.

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  • $\begingroup$ Cute problem. For this discussion maybe alert that it is posed as a problem asking the reader to determine why it is not the counterexample that one wants here although it looks plausible for a moment. $\endgroup$ – B. S. Thomson Jan 1 '16 at 17:26
  • $\begingroup$ Neat example, but because of the discontinuity, it doesn't answer the original question. $\endgroup$ – Ian Jan 1 '16 at 20:50
  • $\begingroup$ @Ian: If I were a detail-oriented person, I would say that this is unclear: OP has still not resolved Michael Hardy's comment regarding the scope of "almost everywhere" in the Question. (This "Answer" was originally posted to highlight this ambiguity. Also, I like it.) $\endgroup$ – Eric Towers Jan 2 '16 at 8:50
  • $\begingroup$ Don't give this to Calculus students haha $\endgroup$ – Math Boy Jan 6 '16 at 4:13

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