How does the Lebesgue measure measure non-cartesian product sets?

Question

Consider the measure space $(\mathbb{R} \times \mathbb{R}, \mathcal{B} \otimes \mathcal{B}, \lambda \otimes \lambda)$ where $\lambda$ is the Lebesgue measure on $(\mathbb{R}, \mathcal{B})$. Since $\lambda$ is $\sigma$-finite, $\lambda \otimes \lambda (B_1 \times B_2) = \lambda(B_1) \cdot \lambda(B_2)$ for $B_1, B_2 \in \mathcal{B}$ (right?).

Then consider the set $\{(x,y) \in \mathbb{R}^2 : x=y\}$. Firstly, how do we know this set is Lebesgue-measurable, and secondly, how would we define the Lebesgue measure of this set? Apparently it is zero, but why?

Answer 1

Let $A=\{(x,y)\in\mathbb{R}^2:x=y\}$. You could use that $B\otimes B$ are the Borel sets on $\mathbb{R}^2$ so a closed subset of $\mathbb{R}^2$ is measurable. If you don't know that yet you can go from first principles: Divide $A$ into a countable union of $A_n = \{(x,y)\in A: -n\leq x\leq n\}$. Show that $A_n$ is measurable by starting with the square $B=[-n,n]\times[-n,n]$ and consecutively cutting out other squares to approximate the line. That way you can also show that the complement of $A_n$ in $B$ has measure $4n^2$ equal to the measure of $B$ and hence $\lambda\times\lambda(A_n)=0$. Since $A$ is the union of such sets it is also of measure $0$.

Answer 2

(I include $0$ in the set N of natural numbers). Let $a_0 =0$ and let $a_n=\sum_{j=1}^{j=n} (1/j)$ for positive integer $ n$ . Observe that $$a_{n+1}<a_n +(1/n) \text { for } n>0$$ and $$k= \sum_{n \in N} (a_{n+1}-a_n)^2 =\sum _{n=1}^{\infty} (1/n^2)< \infty.$$ Now for $ d>0$ let $S(0,d)$ be the interior of the square centered at $(0,0)$ with opposite corners at $(da_1,da_1)=(d,d)$ and $(-da_1,-da_1)=(-d,-d)$. For $n>0$ let $S(n,d)$ be the interior of the square centered at $(da_n,da_n)$ with opposite corners at $(d( a_n -1/n),d(a_n-1/n))$ and $(d(a_n+1/n),d(a_n+1/n))$. For negative integer $ n $ let $S(n,d)=\{(x,y) :(-x,-y) \in S(-n,d) \}$..................... We have now $$\{(x,x) :x \in R\} \subset \bigcup_{n \in Z} S(n,d) =T(d)$$ and $$\lambda (T(d)) \le \sum_{n \in Z} \lambda (S(n,d) =4d^2(1+2k).$$ Since $k$ is fixed (BTW $k= \pi^2/6$), and $d$ can be arbitrarily small, we have $$\lambda (\{(x,x) :x \in R\})=0.$$