Proving $f$ is Riemann integrable without Lebesgue-Riemann criterion.

Asked Apr 25 '21 at 14:19

Active Apr 25 '21 at 14:28

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I have been studying this criterion that tell us that if $f:[a,b]\to \mathbb{R}$ is bounded and almost everywhere continuous respect to $\lambda$ (the Lebesgue measure) iff $f$ is Riemann integrable.

My problem is that I had this example let $f:[0,1]\to \mathbb{R}$

$$f(x)=\left\{\begin{array}{cc} 0& x\in [0,1]\setminus \mathbb{Q}\\ 1/q &\text{ if }p/q \text{ is the lowest expresion for }x\in \mathbb{Q}\cap [0,1] \end{array}\right.$$

So I know this functions is almost everywhere continuous and bounded. So by the criterion is Riemann integrable.

But I wanted to prove that this Riemann integrable with out the criterion, as my intuition tells me that it shouldn't be Riemann integrable but the theorem say it is.

Any help would be greatly apreciated.

edited Apr 25 '21 at 14:28

Bernard

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asked Apr 25 '21 at 14:19

juaaan

Are you familiar with the idea of a uniform convergent series of functions? – GBA Apr 25 '21 at 14:39
@JoséCarlosSantos, yes this answers my question Thank you, – juaaan Apr 25 '21 at 15:13
@gba yes I'm familiar with the uniform convergent series of functions – juaaan Apr 25 '21 at 15:13
There is a nice solution using the fact that if $f_n$ uniformly converges to $f$ and each $f_n$ is integrable, then $f$ is integrable. – GBA Apr 25 '21 at 15:34
@gba that sound really nice, do you know where I can find it? – juaaan Apr 25 '21 at 15:50
I'll post it a little later - at least some main ideas in the proof and I'll let you fill the rest – GBA Apr 25 '21 at 15:59

Proving $f$ is Riemann integrable without Lebesgue-Riemann criterion.

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