The Diagonal in $R^2$ has measure $0$

128 Views Asked by At

Look at $(\mathbb{R^2},B(\mathbb{R^2}),m)$ with $E=\{(x,x)|x\in \mathbb{R}\}$

Prove that $E$ is measurable with $m(E)=0$.

My attempy: I used Tonelli's theorem: If i pick $f=1_E$, I get:

$m(E)=\int_{x\in \mathbb{R}} \int_{y\in \mathbb{R}}1_Edm_ydm_x=\int_{x\in \mathbb{R}} (\int_{y=x}1_Edm_y)dm_x=\int_{x\in \mathbb{R}}0dm_x=0$

Is this correct? Any help proving that E is measurable would be great as well.

1

There are 1 best solutions below

0
On

Your argument is correct. The diagonal is a closed set, hence it is a Borel set.