Let $ f(x,y)=\begin{cases} xy \ \ if\ \ x=y\\ 0 \ \ \ \ if\ \ x\neq y\end{cases}$. Compute $ \displaystyle\int_0^1\int_0^1f(x,y)dxdy $.
I know it's strange to write $\displaystyle\int_0^1\int_0^1f(x,y)dxdy=\int_0^1\int_0^1f(x,x)dxdx$.
Here is what I tried. Let $g(x)=f(x,x)=x^2$. $$ \int_{[0,1]\times [0,1]}f(x,y)dxdy=\int_0^\sqrt{2} g(t)dt$$ (It becomes the integral along the diagonal of the square $[0,1]\times [0,1]$). Does it make sense? Thank you.
By Fubini, this is equal to the double integral over the square; the function is $0$ everywhere except a set of measure $0$, so the integral is $0$.
To see this directly: the inner integral is of a function that is $0$ except at exactly one point, so the inner integral is $0$ for all $y$.