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Everything that follows takes place in the Borel $\sigma$-algebra with Lebesgue measure.

The Lebesgue mean of $f$ at $x$ is defined as $\displaystyle \lim_{\epsilon\to 0} \int_{x-\epsilon}^{x+\epsilon} \frac{f}{2\epsilon}$. The function defined by $[f](x) =$ "Lebesgue mean of $f$ at $x$" is equal to $f$ almost everywhere so integrals of it will be equal to integrals of $f$, call this property $P$.

Starting with the interval $[0,1]$ and removing the middle $1/4$ then the two middle $1/8$-ths then the four middle $1/16$-ths and so on produces the fat Cantor set which has measure $1/2$ but does not contain any interval (Since measures are countably additive this informs us that the set contains uncountably many points). Let $I$ be the indicator function for the fat Cantor set - it is equal to $1$ when applied on a point of the set and $0$ otherwise.

Intuitively I would have thought that the Lebesgue mean $[I]$ of the indicator for the fat Cantor set would be the zero function. Since that contradicts $P$, it seems more plausible that $[I] = I$ but I cannot prove this.

How can we construct the function $[I]$?

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The intervals you remove are open, right? So the point $\frac{3}{8}$ belongs to the fat Cantor set. However $I(\frac{3}{8}+\epsilon)=0$ for $0<\epsilon<\frac{1}{4}$. Hence $[I](\frac{3}{8})\leq\frac{1}{2}$. Still you can have $[I]=I$ almost everywhere. $[I](x)=I(x)$ for $x$ not in the fat Cantor set is easy. –  Rasmus Oct 19 '10 at 19:45
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In the second step you remove two intervals of length 1/16, in the third step, four intervals of length 1/64 etc. –  Florian Oct 19 '10 at 20:23
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The set where the Lebesgue mean is 1 is of first category, so there are many troubles beyond the endpoints of the open intervals. See this open access article by Buczolich for more general results that imply this: projecteuclid.org/…. It would be nice if there were an explicit description of a set of measure 1/2 where the mean is 1, but I don't have any ideas. –  Jonas Meyer Oct 20 '10 at 3:28

1 Answer 1

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The Lebesgue mean does not exist for "boundary points". Let $A_n$ be the union of the intervals of step $n$, that is, $A_1=[0,3/8]\cup [5/8,1]$, $A_2=[0,9/64]\cup [15/64,3/8]\cup [5/8,49/64]\cup [55/64,1]$ etc, $A=\cap_n A_n$ is the fat Cantor set. Consider a boundary point $x$ of a set $A_n$. Consider the sequence $\varepsilon_k:=(3/8)^{n+k}$. Then by symmetry we have $I_k=|(x-\varepsilon_k,x+\varepsilon_k)\cap A|=1/4\cdot(1/2)^{n+k}$ for $k$ large enough, thus $I_k/\varepsilon_k \sim (4/3)^{n+k} \to \infty$.

You can think of elements of $A$ as infinite sequences of {Left,Right}. Those sequences which are eventually constant correspond to "boundary points". My guess is that the points for which the mean is 1 correspond to sequences that are normal in some sense (for example, all patterns of {Left,Right} appear with the same asymptotic frequency).

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