Boundary of volume zero Let L be the boundaryof a rectangle in $$R^2$$ How can I prove that it has volume zero. I have tried to divide it in 4 graphs and show they have olume zero but I haven't got anything
 A: $$\partial([a,b]\times [c,d])\subset [a-\epsilon,b+\epsilon]\times [c-\epsilon,d+\epsilon]\setminus[a+\epsilon,b-\epsilon]\times [c+\epsilon,d-\epsilon]$$
A: The proof will become obvious once you understand what does something has "area/volume/measure" zero means.
For any subset $A$ of $\mathbb{R}^2$, we declare it to has "area/volume/measure" zero if and only if you can cover it by countable many rectangles and the sum of "ordinary" area of the rectangles can be make as small as you like.
Consider the example of a single line segment joining $(0,0)$ and $(1,0)$. 
For any $\epsilon > 0 $, no matter how small $\epsilon$ is, one can cover the line
segment by a rectangle $[0, 1] \times [ -\frac{\epsilon}{2}, \frac{\epsilon}{2} ]$ whose "ordinary" area is $\epsilon$. So the "area/volume/measure" of a line segment is $0$.
Let say you have a rectangle $[a,b] \times [c,d]$. To prove its boundary has "area/volume/measure" $0$, you just need to show for any $\epsilon > 0 $, you can find 4 rectangles covering the four edges of your rectangle and the sum of "ordinary" area of the smaller rectangles is at most $\epsilon$. One choices of the smaller rectangles are:
$$[a,b] \times [ c-\frac{\epsilon}{8(b-a)}, c+\frac{\epsilon}{8(b-a)} ], 
  [a,b] \times [ d-\frac{\epsilon}{8(b-a)}, d+\frac{\epsilon}{8(b-a)} ],\\
  [ a-\frac{\epsilon}{8(d-c)}, a+\frac{\epsilon}{8(d-c)} ] \times [c,d],
  [ b-\frac{\epsilon}{8(d-c)}, b+\frac{\epsilon}{8(d-c)} ] \times [c,d]$$ 
