One special set on [0,1]

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?

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For every $f$, $Y$ can be the set of rational numbers, then the boundary of $A$ has positive measure. –  Did Jun 9 '11 at 5:52
Suppose $f$ is increasing, then $Y$ is of measure zero. –  Asaf Karagila Jun 9 '11 at 5:53
$Y=f^{-1}(0) \cup N$, where $N$ is a set of measure zero. –  Alexander Thumm Jun 9 '11 at 8:41
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). –  Jonas Meyer Jun 12 '11 at 6:45
@all: I agree. Please formulate it as an answer, I will accept. –  Ilya Jun 12 '11 at 6:48