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Let $f(x)$ be a Riemann integrable function on any finite interval $I\subset\mathbb{R}$. Supposing $f(x)=0$ for $x\notin[a,b]$, show that $$\lim\limits_{h\to0}\int_a^b\left|f(x+h)-f(x)\right|\,\mathrm{d}x=0.$$ How to prove that if not familiar with the fact that Riemann integrable function is almost everywhere continuous?

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    $\begingroup$ Are you looking for a proof which do not use measure theory (purely using riemann integral)? $\endgroup$ – user99914 Apr 27 '17 at 10:16
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    $\begingroup$ Yes, by using riemann integral. $\endgroup$ – username Apr 27 '17 at 10:19
  • $\begingroup$ What does "integrable on $\mathbb R$" mean? $\endgroup$ – zhw. Apr 27 '17 at 20:33
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Brutal force: Let $\epsilon >0$. Then since $f$ is Riemann integrable, there is a $\delta >0$ so that any partition $P$ of $$a-1= x_0<x_1 < x_2 < \cdots x_n= b+1$$ with $x_{i+1} - x_i < \delta$ will have

$$ U(f, P)- L(f, P)<\epsilon.$$

Now consider the partition

$$ a < a+ h < a+ 2h < \cdots <a+ kh < b$$

where $a+ (k+1)h \ge b$ and $h < \delta/2$. Then

\begin{equation} \begin{split} \int_a^b |f(x+ h) - f(x)| dx &= \sum_i \int_{a+ ih}^{a+ (i+1)h} |f(x+ h) - f(x)|dx \\ &\le \sum_i h \left( \sup_{x\in [a+ih,a+ (i+2) h]} f(x) - \inf_{x\in [a+ih,a+ (i+2) h]} f(x)\right) \\ &\le U(f, \hat P ) - L(f, \hat P) < \epsilon \end{split} \end{equation}

where $\hat P$ is the partition

$$ a< a+2h< a+ 4h < \cdots < a +\ell (2h) < b,$$

where $a + (\ell +1) 2h \ge b$. Thus

$$\int_a^b |f(x+ h) - f(x)| dx < \epsilon$$

whenever $h < \delta/2$ and we are done.

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  • $\begingroup$ I was half way through my brute force proof, you beat me to it! $\endgroup$ – Harambe Apr 27 '17 at 10:43
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    $\begingroup$ Let's not call this brutal force. It is a wonderful and more natural approach and is way simpler than using measure theory (step functions etc). Simpler is better. +1 $\endgroup$ – Paramanand Singh Apr 27 '17 at 10:51
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    $\begingroup$ Thank you! It helps me a lot.It may be $a+(k+1)h\ge b$? $\endgroup$ – username Apr 27 '17 at 11:25
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Hint.

You can prove the result for continuous functions as a continuous function on a compact interval is uniformly continuous.

Then, if $f$ is Riemann-integrable, you can for all $\epsilon >0$ find a step function $h_\epsilon$ such that: $$ \int_a^b \left\vert f-h_\epsilon \right\vert < \epsilon$$ and finally a function $g_\epsilon$ continuous such that $$ \int_a^b \left\vert h_\epsilon-g_\epsilon \right\vert < \epsilon$$

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  • $\begingroup$ The absolute value should be inside integral. If not, there exist constant function (so trivial step function) such that those integrals have value 0. Anyway I cannot se how those statements are related with question. $\endgroup$ – Przemek Apr 27 '17 at 10:47
  • $\begingroup$ Agreed that the absolute value should be inside the integral. The relationship with the OP can be made by subtracting inside the integral to be minimized a continuous function for which the result is straightforward due to uniform continuity. $\endgroup$ – mathcounterexamples.net Apr 27 '17 at 12:16
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    $\begingroup$ Why not just bypass continuous functions altogether? The result is very simple for step functions, even easier than that for continuous functions. $\endgroup$ – zhw. Apr 27 '17 at 20:43
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Here is a cleaner argument. We know that $C_c(\mathbb{R})$, namely the continuous functions with compact support are dense in $L^1(\mathbb{R})$. With this, denote $f_h \triangleq f(x+h)$. Take a $g \in C_c(\mathbb{R})$ such that $$ ||f-g||_{L^1(\mathbb{R})}<\frac{\epsilon}{2}. $$ Now note by triangle inequality that, $$ ||f_h-f||_{L^1(\mathbb{R})} \leq \underbrace{||f_h-g_h||_{L^1(\mathbb{R})}}_{\leq \frac{\epsilon}{2}} + \underbrace{||g_h-g||_{L^1(\mathbb{R})}}_{\to 0,\ \text{as} \ h\to 0} + \underbrace{||g-f||_{L^1(\mathbb{R})}}_{<\frac{\epsilon}{2}} $$ where the middle convergence follows since, as $|g(x+h)-g(x)|\to 0$ as $h\to0$, due to continuity of $g(\cdot)$ and since $|g(x+h)-g(x)|\leq 2|g(x)|\in L^1$, we have by dominated convergence that $\int |g(x+h)-g(x)|\to 0$.

Hence, for every $\epsilon>0$, $$ \lim_{h\to 0}||f_h-f||_{L^1(\mathbb{R})} < \epsilon. $$ Since $\epsilon>0$ is arbitrary, we conclude.

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