2
$\begingroup$

Let $X = [0,1]$. Define $f:X\to\mathbb{R}_{\geq 0}$ to be Lipschitz continuous on $X$. Put $$Y\subset X:\int\limits_Y f(x)\,dx = 0$$ What can we say then about $A = X\setminus Y$? It is not defined uniquely, but I am interested in some common properties of such sets. Namely, can they be nowhere dense, or with boundary of positive measure?

  • 0
    For every $f$, $Y$ can be the set of rational numbers, then the boundary of $A$ has positive measure.2011-06-09
  • 0
    Suppose $f$ is increasing, then $Y$ is of measure zero.2011-06-09
  • 0
    $Y=f^{-1}(0) \cup N$, where $N$ is a set of measure zero.2011-06-09
  • 0
    If $f$ is an arbitrary nonnegative Lipschitz function, then $Y$ is an arbitrary union of a closed set and a set of measure zero (in the way indicated in Alexander Thumm's comment).2011-06-12

0 Answers 0