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I'm trying to prove that $f(x) = x$ is integrable, using this definition. I've seen somewhere that an equivalent way to prove it is to show that given $\epsilon>0$ we can always have:

$$S(f,P)-s(f,P) <\epsilon$$

Where, duo to the nature of the function $f$, we have:

$$S(f,P) = \sum t_i(t_i-t_{i-1}) = \sum t_i^2-\sum t_i t_{i-1}$$ $$s(f,P) = \sum t_{i-1}(t_i-t_{i-1}) = \sum t_{i-1}t_i-\sum t_{i-1}^2$$

$$S(f,P)-s(f,P) = \sum t_i^2-\sum t_i t_{i-1} -\left(\sum t_{i-1}t_i-\sum t_{i-1}^2\right) =$$ $$\sum t_i^2-2\sum t_it_{i-1}+\sum t_{i-1}^2$$

I don't see how it helps. I know that the sum is in the interval $[0,1]$ so it might help using this. I might take the partition somehow related to the $\epsilon$ so this sum above always end up less than $\epsilon$ but I have no idea of how to do it

Community
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Guerlando OCs
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  • You can define the Riemann-integrability as $\lim_{n\to\infty}S(f,P_n)-s(f,P_n)=0$, where $P_n$ is a partition with $n$ parts. – Masacroso Aug 24 '16 at 22:57
  • "an equivalent way to prove it is to show that given $\epsilon>0$ we can always have: $S(f,P)-s(f,P) <\epsilon$": the true statement is "given $\epsilon>0$, there's a $\delta_{\epsilon}>0$ s.t. for all partition $P$ of mesh $<\delta_{\epsilon}$, we have $S(f,P)-s(f,P) <\epsilon$. – paf Aug 24 '16 at 23:01

2 Answers2

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What you have is $$S(f,P)-s(f,P)=\sum(t_i-t_{i-1})^2.$$ Given $\varepsilon>0$, pick $N$ so that $N>\frac{1}{\varepsilon}$. Now choose the partition $P=\{0,\frac{1}{N},\ldots,\frac{N-1}{N},1\}$. Then we have $$S(f,P)-s(f,P)=\sum\left(\frac{i}{N}-\frac{i-1}{N}\right)^2=\frac{1}{N}<\varepsilon. $$

Aweygan
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Hint: \begin{align*} \sum t_i^2-2\sum t_it_{i-1}+\sum t_{i-1}^2 & = \sum t_i^2-2 t_it_{i-1} + t_{i-1}^2 \\ & = \sum (t_{i} - t_{i-1})^{2} \end{align*} and you can always use the uniform partition

Shai
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