The original form of the question is as follows: Let $f: \mathbb{R}\rightarrow \mathbb{R}$ be a function. Denote $f(x+h)$ as $f_h (x)$. If $f(x)$ is integerable on $[a-1, b+1]$, show that $$\lim_{h\rightarrow 0}\left|\int_a^b f(x)dx-\int_a^b f_h(x)dx\right|=0.$$ My method: $\forall \epsilon>0$, $\forall$partition of $[a,b]$ : $a=x_0<x_1<\ldots<x_n=b$, since $f(x)$ is integrable in $[a,b]$, by Lebesgue’s criterion, for each interval $[x_i, x_{i+1}]$, $\exists \eta_i\in (x_i, x_{i+1})$ such that $f$ is continuous at $\eta_i$. For each $i$, let $h_i>0$ be a real number such that $(\eta_i-h_i, \eta_i+h_i)\subset [x_i, x_{i+1}]$ and $(|x-\eta_i|<h_i)\Rightarrow (|f(x)-f(\eta_i)|<\frac{\epsilon}{b-a})$. Then $$h<\min\{h_0, h_1,..., h_{n-1}\}\Rightarrow \sum_{i=0}^{n-1}|f_h(\eta_i)-f(\eta_i)|\Delta x_i<\epsilon.$$ I want to know if this can show what we want. To promote it into a more general case, suppose $f_h(x): \mathbb{R}\rightarrow\mathbb{R}$ is a function with a parameter $h$, and it is integrable on $[a,b]$ when $0<|h|<1$. If $\forall \epsilon>0$ $\forall$ partition, $\exists \delta>0$ such that $\forall h$ such that $0<|h|<\delta$$\exists$ a series of signal points, with their Riemann sum (The signal points can vary while $h$ varies.)with respect to the partion, the value of which is denoted as $S$, such that $|S|<\epsilon$, then can it be inferred that $\lim_{h\rightarrow0}\int_a^bf_h(x)dx=0$? Prove or give counterexamples.
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https://math.stackexchange.com/questions/1018716/translation-operator-and-continuity – Evangelopoulos Foivos Mar 11 '22 at 07:27
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thanks, but I don't think this answers if my promotion of the question is right – Asigan Mar 11 '22 at 08:12
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1I am not sure I get your "promotion". For instance, what if $f_h = f \equiv 1$ for all $h$? – hal4math Mar 11 '22 at 11:14
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Well, but for each $h$, the Riemann sum of any partition for $[a, b]$ is $b-a>\frac{b-a}{2}$, hence it does not fit the condition of my 'promotion'. Furthermore, I would like to know where my description is ambiguous or confusing, and I will edit my question. @hal4math – Asigan Mar 11 '22 at 12:53
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I see. Well, for instance I don't follow the relationship between $\delta$, $h$ and $\varepsilon$. Specifically, how does the $\delta$ relate to the $\varepsilon$ and the partition? – hal4math Mar 11 '22 at 13:13
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You can choose arbitrary $\epsilon$ and the partition. Then a $\delta$ can be found, but different $\epsilon$ and partitions can have different $\delta$. Then $\forall h$ such that $0<h<\epsilon$, for example, for $h=\frac{\delta}{2}$ you can choose a series of signal points with respect to the partition, such that its Riemann sum is smaller than $\epsilon$. And for different $h$ the signal points can vary. (My question is indeed ambiguous here. I will edit it.) – Asigan Mar 11 '22 at 13:28
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1@Asigan: the firs part can be simplified greatly. For example assume $h>0$ for simplicity. Then $$\begin{align} \int^b_af(x),dx-\int^b_af(x+h),dx&=\int^b_a f(x),dx- \int^{b+h}_{a+h}f(x),dx\ &=\int^{a+h}_af(x),dx -\int^{b+h}_bf(x),dx \end{align}$$ Being $f$ Riemann integrable over $[a-1,b+1]$, $f$ is bounded in that interval. Say $|f(x)|\leq M$ for all $x\in[a-1,a+1]$. Then $$\begin{align} \Big|\int^{a+h}_af-\int^{b+h}_bf\Big|&\leq \int^{a+h}_a|f|+\int^{b+h}_b|f|\leq 2Mh \end{align}$$ – Mittens Mar 11 '22 at 14:18
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My intuition tells me that there should be a counterexample. Because I think your "promotion" is something similar to switching the order of the net limit of the Riemann integral and the function limit for $h\to 0$. That usually needs quite strong properties. – hal4math Mar 11 '22 at 14:21
1 Answers
My vote for the best answer goes to the ever reliable Oliver Diaz who shows that you do not need any sophisticated property of the Riemann integral, nor do you need to muck around in the definition with exotic computations involving Riemann sums. [Moral: never return to the actual definition of the Riemann integral except when forced at gunpoint.]
You need to observe, says Oliver, only that in the case $h>0$:
$$\begin{align} \int^b_af(x)\,dx-\int^b_af(x+h)\,dx&=\int^b_a f(x)\,dx- \int^{b+h}_{a+h}f(x)\,dx\\ &=\int^{a+h}_af(x)\,dx -\int^{b+h}_bf(x)\,dx \end{align}.$$
But there is always an opportunity to learn other methods. In fact who cares about solving problems? The only point is to acquire methods.
Here is a theorem every student of the Riemann integral should learn.
Theorem (Arzelà-Osgood) If $\{f_n\}$ is a uniformly bounded sequence of Riemann integrable functions that converges pointwise to another Riemann integrable function $f$ on an interval $[a,b]$, then $$ \lim_{n\to \infty} \int_a^b f_n(x)\,dx = \int_a^b f (x)\,dx.$$
Advanced students recognize that as the precursor to Lebesgue's bounded convergence theorem.
In this case take any sequence of real numbers $\{h_n\}$ convering to zero. For every such sequence define $f_n(x)=f(x+h_n)$. Observe that this sequence $\{f_n\}$ is uniformly bounded and converges pointwise to $f$. Apply the Arzelà-Osgood theorem: $$ \lim_{n\to \infty} \int_a^b f(x+h_n)\,dx = \lim_{n\to \infty} \int_a^b f_n(x)\,dx =\int_a^b f (x)\,dx.$$
Since this is true for any such sequence it is also true for $h\to 0$.
While this is like using a sledge hammer as a nutcracker, mathematics students should get used to carrying around heavy duty equipment. It is more fun to smash a problem to bits rather than fuss with $\epsilon$'s and $\delta$'s. Also never treat the Riemann integral with undue respect: it is just a teaching tool.
REFERENCES:
[1] Arzelà , C. (1885). Sulla integrazione per serie. Atti Acc. Lincei Rend. 1(4): 532–537 and 596–599.
[2] Osgood, W. F. (1897). Non-uniform convergence and the integration of series term by term. Amer. J. Math. 19: 155–190.
[3] Lewin, J. W. (1986). A truly elementary approach to the bounded convergence theorem. Amer. Math. Monthly. 93(5): 395–397.
[4] Luxemburg, W. A. J. (1971). Arzelà’s dominated convergence theorem for the Riemann Integral. Amer. Math. Monthly. 78(9): 970–979.
[5] Kestelman, K. (1970). Classroom Notes: Riemann integration of limit functions. Amer. Math. Monthly. 77(2): 182–187.
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Thanks for your answer. But I am now more curious for my promotion. Could you please prove it or give a counterexample? – Asigan Mar 12 '22 at 15:51
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I don't see how your 'promotion' using Riemann sums is worth pursuing. If you want some kind of limit like $\lim_{h\rightarrow0}\int_a^bf_h(x)dx$ you can handle it by just considering sequences $h_n\to 0$ and use then standard convergence theorems. Those are the most imporant tools and, at best, you would just be reproving them by fussing with Riemann sums. – B. S. Thomson Mar 12 '22 at 17:50
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