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?
One special set on [0,1]
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real-analysis
general-topology
functional-analysis
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0For every $f$, $Y$ can be the set of rational numbers, then the boundary of $A$ has positive measure. – 2011-06-09
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0Suppose $f$ is increasing, then $Y$ is of measure zero. – 2011-06-09
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0$Y=f^{-1}(0) \cup N$, where $N$ is a set of measure zero. – 2011-06-09
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0If $f$ 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