# Conditions for applying the second fundamental theorem of calculus with gauge integrals

I was thinking about this question while walking home today and can't seem to prove or come up with a counterexample myself.

Let $$f:[a,b]\rightarrow\mathbb{R}$$ be a continuous function, $$f(x),$$ differentiable everywhere on $$(a,b)$$, except for one point $$\xi$$ in $$(a,b)$$ where $$f'(\xi)$$ does not exist. Since $$f$$ is continuous everywhere and differentiable everywhere on $$(a,b)$$ except at $$\xi$$, $$f'$$ is gauge integrable on $$[a,b]$$. Am I correct that $$\int_a^bf'(x)\;\mathrm{d}x=f(b)-f(a)$$ despite that nasty point $$\xi$$ in $$(a,b)$$? And could this value be interpreted as the "signed" area under the curve from high school calculus?

• What makes you thing $f'$ is integrable? – Kavi Rama Murthy Dec 13 '18 at 23:52
• I thought that $f'$ is gauge integrable because $f$ is differentiable everywhere except at a countable number of points? – user626213 Dec 14 '18 at 0:42
• I do not know what 'gauge integrable' means but I think $f'$ need not be integrable in any sense. Note that $f'$ need not be bounded. – Kavi Rama Murthy Dec 14 '18 at 5:32

If $$f(x)=x\sin (\frac 1 x)$$ ($$f(0)=0$$) then $$f$$ satisfies your hypothesis but $$f$$ does not satisfy the Fundamental Theorem of Calculus . If it did satisfy this on each interval contained in $$[-1,1]$$ the it would be absolutely continuous and hence of bounded variation on $$[-1,1]$$ but it is not of bounded variation.