Double Integrals — Question 9

Question 9 – Solution

Strategy. The diagonal has no area: trap every possible nonzero upper-sum contribution in a shrinking collection of grid squares.

Step 1: Lower sums Every subrectangle with positive area contains a point off the diagonal, where $f=0$. Its infimum is therefore $0$, so every lower sum equals $0$.

Step 2: Upper-sum cover In an $n\times n$ grid, each square has area $1/n^2$. The $n$ diagonal squares meet $x=y$. At each of the $n-1$ interior diagonal grid vertices, two additional off-diagonal squares have closures meeting the diagonal. Thus at most $$n+2(n-1)=3n-2$$ closed grid squares have supremum $1$; all others have supremum $0$. Consequently, $$0\le U_n\le\frac{3n-2}{n^2}<\frac 3n.$$

Step 3: Integrability Given $\varepsilon>0$, choose $n>3/\varepsilon$. Then $$0\le U_n-L_n<\frac 3n<\varepsilon.$$ The Darboux criterion proves integrability, and the integral is squeezed between zero lower sums and upper sums tending to zero: $$\boxed{\iint_{[0,1]^2}f\,dA=0}.$$ The function is discontinuous at every point of the diagonal, but that one-dimensional set contributes no area.