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 $[I]$?