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if $f$ is integrable, i wish to show

$\frac{n}{2} \int_{-1/n}^{1/n} (f(x+y) - f(x))dy \to 0 $ as $ n \to \infty $

this looks to be very intuitive, but im having trouble proving it formally. any tips?

EDIT: added the n infront of the integral, thanks Jonas

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The question doesn't really make sense as stated. Do you mean $n\int_{-1/n}^{1/n}f(x+y)dy\to 2f(x)$? This will not hold in general, but you may want to look at Lebesgue points. en.wikipedia.org/wiki/Lebesgue_point –  Jonas Meyer Mar 17 '11 at 21:43
    
You have not fixed the question. You still probably want an $n$ in front of the integral. If this is really what you want to ask, then the answer is yes, because $\int_{-1/n}^{1/n}f(x)dy\to0$ (trivially) and $\int_{-1/n}^{1/n}f(x+y)dy\to0$ (because $f$ is integrable, for all $\epsilon>0$ there exists $\delta>0$ such that if the measure of $A$ is less than $\delta$ then $|\int_A f|<\epsilon$). –  Jonas Meyer Mar 17 '11 at 22:20

1 Answer 1

up vote 3 down vote accepted

A plausible interpretation of the question is that you're asking whether $\int_{-1/n}^{1/n}f(x+y)dy$ is asymptotically equivalent to $\frac{2}{n}f(x)$ as $n\to\infty$, which amounts to the same as asking whether $\frac{n}{2}\int_{-1/n}^{1/n}f(x+y)dy\to f(x)$ as $n\to\infty$. If $f$ is locally integrable, then this holds almost everywhere by Lebesgue's differentiation theorem. You could not hope for more than this, because changing the value of $f$ at $x$ (or on any set of measure $0$) will not affect the integrals.

Edit: Since I first posted, the question was edited to what I answered.

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