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Let be $f:A\longrightarrow\mathbb{R}$ a any function, defined in block $A$. If there is a limit $\lim_{|P|\rightarrow 0} \sum(f;P*)$ then f is limited.

* My proof: Suppose $f$ unlimited, by contradiction. Then take a sequence of blocks $B_1, ... , B_k \in P$. Hence, since we suppose $f$ unlimited there exists at least one $B_{\xi}$ such that $|f(x_{\xi})|\ge M$, where $M$ any constant. So we have to $\sum(f;P*)=\sum_{i=1, i\ne \xi}^k(f(x_i)vol.B_i) + f(x_{\xi})vol.B_{\xi}$. Then, as the second portion diverges and the first converges, then the sum diverges. Thus we get the result.

Sorry for my bad English. I gave a demonstration to this problem but I discovered a mistake, so I decided to ask for help for you.

Thanks in advance.

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    If possible, we would like you to post your demonstration here, with the mistake. It will help us figure out why you made the mistake, and provide context for your question. – 2017-01-25
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    I just put the demo. – 2017-01-26
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    Ok, so what do you think is the mistake? And +1 for putting up the proof. – 2017-01-26
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    I saw that, the function can compensate for the part that is unlimited. Thus making Riemann's sum do not differ as I want it to. – 2017-01-26
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    Note that $vol(B_{\xi})$ may go to zero as $f(x_{\xi})$ diverges. – 2017-01-26
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    If $f$ is unbounded, then if it is unbounded on some block $B$ with non-zero volume, then consider that block $B$ for the partition. – 2017-01-26
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    What is a block? A hyper rectangle $(a_1,b_1)\times\dotsb(a_n,b_n)$ in $\mathbb R^n$ with finite volume? – 2017-01-26
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    I did not understand. Do I have to take a covenient partition? And why does vol tend to zero if f is unlimited? The volume should grow absurdly too, right? – 2017-01-26
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    If we assume that f is unlimited only the problem is solved. But if the function is also unlimited inferiorly, then it may occur that the sum of riemann converges, canceling out the plots in which the function diverges. – 2017-01-26
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    What's up? Someone knows how to solve this problem. I need a lot to solve I actually have a list to deal with until Monday. – 2017-01-26

1 Answers 1

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You need to show by contradiction that if $f$ is unbounded in $A$, then it is impossible for the limit of Riemann sums, $\lim_{|P| \to 0} S(f,P)$, to attain a finite limit.

Assume there exists $\epsilon_0 >0$ such that for any number $J \in \mathbb{R}$ and any $\delta >0$ there exists a partition $P$ with $|P| < \delta$ and

$$|S(P,f) - J| =|\sum_{i \neq \xi} f(x_i) vol(B_i) + f(x_\xi)vol(B_\xi) - J| > \epsilon _0.$$

Just choose $\epsilon_0 = 1$ and given $\delta > 0$ choose any partition with $|P | < \delta$. As you stated, there exists at least one index $\xi$ such that $f$ is unbounded on $B_\epsilon$.

Now select $x_\xi \in B_\xi$ such that

$$|f(x_\xi)| > \frac{1 + | J - \sum_{i \neq \xi} f(x_i) vol(B_i)|}{vol(B_\xi)}.$$

Using the reverse triangle inequality we have

$$|\sum_{i \neq \xi} f(x_i) vol(B_i) + f(x_\xi)vol(B_\xi) - J| \geqslant |f(x_\xi)|vol(B_\xi) - |J - \sum_{i \neq \xi} f(x_i) vol(B_i)| > 1.$$

This contradicts the existence of the limit so we must have $f$ bounded.

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    Thanks. I realized the error of my demonstration. I should have chosen this J to be the limit and used the definition in my favor. – 2017-01-26
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    @LucianoGauss: You're welcome. Your comment summarizes this well. Only considering the sum is insufficient since, as stated in an earlier comment, the volume of the sub-rectangles can be arbitrarily small. – 2017-01-26
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    I understood. The difficult thing about this demonstration is to think of the number $ 1 + | J-Σi ≠ ξf (xi) vol (Bi) | vol (Bξ) $. But with some specific accounts I could see where it came from. – 2017-01-26