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I'd really appreciate your help with the following problem:

Let $f$ be a integrable function in $[a-1,b+1]$. I need to prove that: $$\lim_{h\to 0} \int_{a}^{b} |f(x+h)-f(x)|dx=0 .$$

Basically I need to show that $\forall \varepsilon \exists \delta .|h|< \delta \to \left|\int_{a}^{b}|f(x+h)-f(x)|dx\right| < \varepsilon $.

In addition, I know that because of $f$ being integrable I can write that $\sum \limits_{i=0}^{n-1}(\sup f-\inf f) \Lambda x_{i} < \varepsilon$ for a division of $[a,b]$.

I tried to define $g=|f(x+h)-f(x)|$ and work with that, but it didn't lead me to any smart conclusion. Also I tried to put the limit inside of the integral and to use the derivative, but I'm not sure when I can do that.

Thank you for the help!

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What is $\Lambda$? –  Gerry Myerson Nov 25 '11 at 9:57
    
$\Lambda x_{i}$-It's the i- interval in the division of [a,b]. –  Jozef Nov 25 '11 at 10:05
    
Is $\Lambda x_i$ similar or the same as $\Delta x_i$? –  robjohn Nov 25 '11 at 10:18
    
It's the same..Sorry for the misunderstanding. –  Jozef Nov 25 '11 at 10:22
    
So how do you multiply an interval by a number, and get a number? Maybe you mean $\Lambda x_i$ is the length of the $i$th interval in the division. –  Gerry Myerson Nov 25 '11 at 10:27

2 Answers 2

up vote 5 down vote accepted

This can be proven by approximating $f$ by a continuous function $g$ so that $\int_a^b|f(x)-g(x)|\mathrm{d}x<\epsilon$. Since $g$ is continuous on a compact set, it is uniformly continuous, so there is a $\delta>0$ so that $|g(x+\delta)-g(x)|<\frac{\epsilon}{b-a}$. Therefore, $$ \begin{align} \int_a^b|f(x+\delta)-f(x)|\mathrm{d}x &\le\int_a^b|f(x+\delta)-g(x+\delta)|\mathrm{d}x\\ &\qquad+\int_a^b|g(x+\delta)-g(x)|\mathrm{d}x\\ &\qquad+\int_a^b|f(x)-g(x)|\mathrm{d}x\\ &\le\epsilon+\frac{\epsilon}{b-a}(b-a)+\epsilon\\ &=3\epsilon \end{align} $$

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who is this $g$ basically? why is it continuous? and why $\int_a^b|f(x)-g(x)|\mathrm{d}x<\epsilon$? –  Jozef Feb 9 '12 at 12:35
    
I was using the fact that the continuous functions are dense in $L^1$. How this is shown depends on whether we are using the Lebesgue or Riemann integral. –  robjohn Feb 9 '12 at 23:21

Let $\varepsilon > 0$ and take a partition of $[a-1,b+1]$ for which $\sum (\sup f - \inf f)\Lambda x_i < \varepsilon(b-a)$.

Now, since $f$ is integrable in $[a-1,b+1]$ let $M$ be the essential supremum of $f$ there (assuming we're talking about proper integrals). You get that any $|h|<1$ satisfies that $|f(x+h)-f(x)|<2M$.

Take $\delta>0$ such that $\delta$ is less than the minimal length of an interval in the partition and less than $\varepsilon$, for a given interval $I$ take the part of $[a,b]$ where $x\in I$ yet $x+h \not \in I$, this has length of at most $\delta$, which is smaller than $\varepsilon$, and the same goes for tha part of $[a,b]$ where $x\notin I$ yet $x+h \in I$. You get that integrating over all of these parts is bound from above by $4Mn\varepsilon$ where $n$ is the number of intervals in the partition. Integrating over all the rest is bound by $2\varepsilon(b-a)$.

All in all the value of the integral is bound by a function of $\varepsilon$, which concludes the proof.

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Can you please explain: "for a given interval $I$ take the part of $[a,b]$ where $x\in I$ yet $x+h \not \in I$"? Toda Thanks! –  Jozef Nov 25 '11 at 10:55
    
Your partition is actually a set of disjoint intervals covering $[a-1,b+1]$. When you take one of the intervals in the partition (say, $[c,d]\subset[a-1,b+1]$), then for some values of $x$ you get that $x\in[c,d]$ yet $x+h$ isn't (for example, $x=b-h/2$, when $0<h<(b-a)$. Then take this entire set... –  Shai Deshe Nov 25 '11 at 11:00
    
It appears that $\delta$ is dependent on the partition used. Don't we need to find a $\delta$ that is independent of partitiion? –  robjohn Nov 25 '11 at 14:35
    
No, since the partition depends on epsilon, and delta may depend on epsilon as well. –  Shai Deshe Nov 26 '11 at 21:59

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