Let $\mathcal{L}$ and $\mu$ denote the Lebesgue measure and counting measures on $[0,1]$, respectively. Prove that the diagonal $D=\{(x,x):x\in[0,1]\}$ is not a $\sigma$-finite set with respect to $\mathcal{L}\times\mu$.
Attempt: Suppose $D$ is $\sigma$-finite set. Then $D=\bigcup_{i=1}^\infty R_i$ for some $\mathcal{L}\times\mu$-measurable sets $R_1, R_2, ... $ with finite measure. So, for any $i\in\mathbb{N}$ and any $x\in [0,1]$, the $x$-section $(R_i)_x:=\{y\in[0,1]:(x,y)\in R_i\}$ has finite $\mu$-measure. So $sup_{x\in [0,1]}\{\mu((R_i)_x)\}$ exists for all $i$. Now, for any $i\in\mathbb{N}$, we choose $x_i\in[0,1]$ such that $\mu((R_i)_{x_i})=sup_{x\in [0,1]}\{\mu((R_i)_x)\}$. Then $[0,1]=\bigcup_{i=1}^{\infty}(R_i)_{x_i}$ Note that $(R_i)_{x_i}$ is finite for any $i$. It follows that $[0,1]$ is countable, contradiction. But I didn't use any Lebesgue measurability.
Second approach: Suppose $D$ is $\sigma$-finite set. Then $D=\bigcup_{i=1}^\infty A_i$ for some $\mathcal{L}\times\mu$-measurable sets $A_1, A_2, ... $ with finite measure. Then there is a $\mathcal{L}\times\mu$-measurable rectangle $R_i\supseteq A_i$ such that $R_i=[b_i,c_i]\times S_i$ has finite $\mathcal{L}\times\mu$-measure for all i. It follows that $[0,1]=\bigcup_{i=1}^\infty S_i$. But since $\infty>\mathcal{L}\times\mu(R_i)=\mathcal{L}([b_i,c_i])\mu(S_i)$, we have $\mu(S_i)<\infty$ i.e $S_i$ is a finite set for all $i$, contradiction.
Can anyone check my attempts? Thanks.