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The name explains it all. I searched for it in MSE and came across a similar [one] but more simpler1. I was interested to know if we can prove that, i.e., given an arbitrary shape (closed and continuous) with an integer area, can we can always divide it as chunks which can have any shape but have unit area only. Also, if such a division is possible, is it unique?

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There is no way it would be unique, just shift some subchunks of each of the chunks around. On the other hand, if we are keeping track of the total area by an integral that sweeps across the real line, since it's a continuous function, we can stop at each integer and create a division – muzzlator Mar 28 '13 at 15:45
up vote 3 down vote accepted

Pick any direction in the plane you like. Vertical is an easy one. Now let a vertical line move in from $-\infty$ until there is one unit of area to its left, calling the value $x_1$. We would express it as $\int_{-\infty}^{x_1}y\ dx = 1$ There is your first piece. Then keep going, find $x_2$ such that $\int_{x_1}^{x_2}y\ dx = 1$ and so on.

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really clever!!!! – dineshdileep Mar 28 '13 at 16:37
@dineshdileep: for reasonable shapes you can modify this in various ways. Having bitten off an area of 1, you can pivot around the intersection of the line with the boundary until you get a "pie-slice" of unit area, then keep going. – Ross Millikan Mar 28 '13 at 16:42

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