Determining null sets with Tonelli's theorem How can I show that the diagonal $D=\{(x,y)\in\mathbb R^2\vert  x=y\}$ is a Lebesgue-nullset in $\mathbb R^2$ by utilizing the theorem of Tonelli?
My solution so far, but it doesn't seem quite right:
$\int_D 1d\lambda = \int_{-\infty}^\infty\int_{y}^{y}1dxdy=\int_{-\infty}^\infty [x]^y_ydy=\int_{-\infty}^\infty0 dy=0$
Can I do it this way?
 A: @Sigma, Your approach is essentially correct. You can prove the diagonal $D=\{(x,y)\in\mathbb R^2\vert  x=y\}$ is a nullset directly from Tonelli's theorem. Here are the details.
Since $D$ is closed in $\mathbb R^2$, $D$ is measurable and so $1_D$ is a non-negative measurable function. So we can apply Tonelli's theorem.
\begin{align*} \lambda(D)&=\int_{\mathbb R^2} 1_D d\lambda = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}1_D(x,y) dx dy= \\
&=\int_{-\infty}^{+\infty}\left(\int_{-\infty}^{+\infty}1_D(x,y) dx\right) dy = \\
&=\int_{-\infty}^{+\infty}\left(\int_{-\infty}^{+\infty}1_{\{y\}} dx\right) dy = \\
&=\int_{-\infty}^{+\infty} 0\; dy =0
\end{align*}
A: A possible approach: first rotate $D$ to the horizontal axis, and recall that the Lebesgue measure is invariant under rotations. Then observe that
$$
(-\infty,+\infty) \times \{0\} = \bigcup_{n \in \mathbb{N}} \left( [-n,n] \times \{0\} \right).
$$
If you prove that $\mathcal{L}^1 \left( [-n,n] \times \{0\} \right)=0$ for every $n$, then you can use the subadditivity of the measure to conclude.
We must show that $\mathcal{L}^1 \left( [-n,n] \times \{0\} \right)=0$ for every $n$. Pick $\varepsilon >0$ and consider the set
$$
\Omega_\varepsilon = [-n,n] \times (-\varepsilon,\varepsilon).
$$
Compute $\int_{\Omega_\varepsilon} \mathrm{d}x \, \mathrm{d}y$ and prove that the result is of order $\varepsilon$. Since $\varepsilon>0$ is arbitrary...
Edit: the first step is only for convenience. One can also consider the set $\Omega_\varepsilon$ lying between the lines $y=x+\varepsilon$ and $y=x-\varepsilon$.
