Fubini's Theorem: Integral Existence for Non-Continuous Functions

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hello,
I posted a question a while ago about fubini's theorem, and i believe i have found my answer.. but to clear things up, i have one more question.

If a function f(x) is not continuous then the integral (by fubinis) does not exist.
is this correct. ( I believe it is correct, but i would feel better if someone else agreed as well)
thank you!
adam
 
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IIRC, a (dumbed down) version of Fubini in analysis says that the order of integration in a double integral may be interchanged if the integrand is a continuous function. That doesn't mean the integral doesn't exist when if the integrand is not continuous.

The LateX images in your other post are not showing, so we can't help you there.
 
A function is integrable iff the set of discontinuities is a measure-0 set. Fubini's theorem has little to do with this - it's about double integration.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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