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In the proof of the theorem, it is supposed, that for a function, which is monotone and integrable on the interval $[a,b],\ f(a) < f(b)$, and $\delta$ is chosen to be equal to $\frac{1}{2}\frac{\epsilon}{f(b)-f(a)}$

Then the question is whether $\sum_{i=1}^n (w_i(f)\Delta x_i )< \epsilon$,

where $m_i=\min f(x)=f(x_{i-1})$ and $M_i=\sup f(x)=\max f(x)=f(x_i)$ so that $w_i=M_i-m_i=f(x_i)-f(x_{i-1})$. In both cases, $x$ belongs to $[x_{i-1},x_i]$.

So, $\sum\limits_{i=1}^nw_i(f)\cdot\Delta x_i=\sum\limits_{i=1}^n(f(x_{i})-f(x_{i-1})\Delta x_i=\sum\limits_{i=1}^n(f(x_i)-f(x_{i-1})\delta$

My question is why $\Delta x_i$ is substituted with $\delta$ if the expression that $\delta$ concerns values of the given argument, not intervals of $x$. Please clarify if you can!

cmk
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user
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1 Answers1

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$\Delta x_i$ is chosen to be $\delta$ (or actually, anything not larger than $\delta$), so that the final sum can be estimated by $$\delta\sum_{i=1}^nf(x_i)-f(x_{i-1})=\delta(f(x_n)-f(x_0))\leq \delta (f(b)-f(a))=\frac{\epsilon}{2}$$ by the definition of $\delta$.

uniquesolution
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  • Yes, but how can $\delta x_i$ be chosen to be $\delta$? – user May 28 '19 at 19:49
  • I guess that depends on the context of the theorem. Usually you want to prove that a certain partition of $[a,b]$ exists, and so you are free to choose the points $x_i$ as you wish, but a precise answer requires the full statement of the theorem. – uniquesolution May 28 '19 at 19:51