Is $(x,x)$ where $x\in A\subset \mathbb{R}$ measurable? 
Let $A\subset \mathbb{R}$ and let $D\subset \mathbb{R^2}$ be defined
$$D=\{(x,x):x\in A\}$$
Show that $D$ is Lebesgue measurable and evaluate $m(D)$

I have few contradictions thoughts on how to tackle this question.
Intuitivly $D$ is a straight line in the $\mathbb{R^2}$ and we know that $\mathbb{R}$ is Lebesgue measure and $m(\mathbb{R})=\infty$ so it is "equivalent" for finding $m(\mathbb{R})=\infty$ where $\mathbb{R}\subset \mathbb{R}^2$
On the other $(x,x)$ is can de defined as a fixed point function $f(x)=x$ so maybe we can use prove the $f$ is measurable function and then prove that both the domain and codomain are measurable
Any suggestion how should I tackle this?
 A: You know that Lebesgue measurable sets are a complete measure space, i.e. if $N \subset P$ where $P$ is measurable with $m(P)=0$ than also $N$ is measurable and $m(N)=0$.
Due to the fact that $D \subset \{(x,y) \in \mathbb{R}^2 : x=y\}$ and the last one is a subspace of dimension 1 of $\mathbb{R}^2$ we can conclude that $D$ is measurable and $m(D)=0$.
A: It has Lebesgue measure 0.
For any $A \subset \mathbb{R}$ we have that $D(A) \subset D(\mathbb{R})$ where $D(A) \overset{\text{def}}{=} \{ (x,x) \ | \ x \in A\}$. 
It is pretty standard (first few pages of any textbook) that if $\lambda $ is the Lebesgue measure then 


*

*$\lambda (\mathcal{B}_{\epsilon}(x)) = \epsilon$, i.e. the measure of an $\epsilon$-ball is $\epsilon$

*$ (\forall i,j) (A_i \cap A_j = \emptyset) \implies \lambda(\cup_{i=1}^\infty A_i) = \sum_{i=1}^\infty\lambda( A_i)$, i.e. $\lambda$ is countably additive


Take any line segment $\mathcal{L}$ of $D(\mathbb{R}) $ of length $l$. It is easy to see that if we divide $\mathcal{L}_l$ into $n$ smaller segments and cover them by $n$ disjoint $\epsilon$-balls for $\epsilon = \frac{l}{2n}$ then the area of the region is $\sum_{i=1}^{n} \pi (\frac{l}{2n})^2 = \frac{\pi}{n}(\frac{l}{2})^2$. Using properties 1 and 2 above we see that $\lambda (\mathcal{L}_l) \leq \lim_{n \to \infty}\frac{\pi}{n}(\frac{l}{2})^2= 0 $. 
Finally, notice that $D(\mathbb{R}) = \cup_{i = 1}^{\infty} \mathcal{L}_i$ for some disjoint line segments $\mathcal{L}_i$ so that by property 2 above we have that $\lambda (D(\mathbb{R}))  =   \lambda (\cup_{i = 1}^{\infty} \mathcal{L}_i) = \sum_{i=1}^\infty\lambda (\mathcal{L}_i)=0$
Therefore, once again  by property 2, we have that $D(A) \subset D(\mathbb{R}) \implies \lambda (D(A)) \leq   \lambda (D(\mathbb{R})) \implies \lambda (D(A)) = 0 $. 
