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.

  • 1
    $\begingroup$ What is the measure of the domain on which $f\ne 0$? $\endgroup$ – Mark Viola Oct 3 '16 at 21:18
  • $\begingroup$ $f$ is almost everywhere $0$. $\endgroup$ – user251257 Oct 3 '16 at 21:19
  • $\begingroup$ $f(x,y)=0$ since the line segment $\{(x,x): 0\leqslant x\leqslant 1\}$ is the image of a subset of $\mathbb R^1$ under a Lipschitz continuous map. $\endgroup$ – Math1000 Oct 3 '16 at 21:21
  • $\begingroup$ Thank you so much. I wanted to compute it directly :-) But Using the fact that "If $ f=0 $ a.e. then $\int f =0 $", we see that the above integral is 0. $\endgroup$ – Jax Oct 3 '16 at 21:29

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 $y$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.