Re: Lebesgue measure on RxR

(i) You computed the measure of the triangle whose vertexes are $\displaystyle (0,0)$, $\displaystyle (0,c)$, $\displaystyle (c,0)$, but it's not what it's asked. You can write $\displaystyle \left\{(x,y)\in\mathbb R^2, x+y=c\right\} =\bigcup_{n\in\mathbb Z}\left\{(x,y)\in\mathbb R^2, x+y=c, n\leq x\leq n+1\right\}$ and show that for all $\displaystyle n$, $\displaystyle m_2(\left\{(x,y)\in\mathbb R^2, x+y=c, n\leq x\leq n+1\right\})=0$. To do that, you can cover the line for a fixed $\displaystyle k\in\mathbb N$ by $\displaystyle k$ squares with measure $\displaystyle \frac 1{k^2}$.

(ii) Write $\displaystyle A = \left[0,\frac{\pi}3\right]\times B$ where $\displaystyle B =\left\{y\in\mathbb R, 0\leq y\leq \frac{\pi}2, \sin y \notin \mathbb Q\right\}$. We can compute the measure of the complement of $\displaystyle B$ in $\displaystyle \left[0,\frac{\pi}2\right]$ writing $\displaystyle \left[0,\frac{\pi}2\right]\setminus B = \bigcup_{r\in \mathbb Q}\left\{x\in \left[0,\frac{\pi}2\right],\sin x=r\right\}$.

Re: Lebesgue measure on RxR

(i) Oh, I see. And then one lets $\displaystyle k \rightarrow \infty$ and we obtain $\displaystyle m_2(\{(x,y)\in\mathbb{R}^2; x+y=c, n\le x\le n+1\})=0$ and since the original set is a disjoint union of sets of measure zero we get at it is also of measure zero.

(ii) Here I fully understand what you mean but I'm not sure of my result.

So what I get is $\displaystyle m_2([0,\pi/2]\setminus B) = \sum_{r\in\mathbb{Q}}m(\{y\in\mathbb{R}; 0\le y\le \pi/2, \sin(y)=r\})=0$. And then $\displaystyle m([0,\pi/2])=m([0,\pi/2]\setminus B\cup B) = m([0,\pi/2]\setminus B)+m(B)=m(B)$. So $\displaystyle m_2([0,\pi/3]\times B)=m([0,\pi/3])m(B)=(\pi/3)(\pi/2)=\pi^2/6$.

Is this correct? What I'm not sure about is the step $\displaystyle m_2([0,\pi/3]\times B)=m([0,\pi/3])m(B)$, from what I've read the set needs to be a "rectangle" for that to be allowed.

Re: Lebesgue measure on RxR

(i) It's ok.

(ii) We can write, since the measures of the following sets are finite: $\displaystyle m_2\left(\left[0,\frac{\pi}3\right]\times B\right) =m_2\left(\left[0,\frac{\pi}3\right]\times \left[0,\frac{\pi}2\right]\right)- m_2\left(\left[0,\frac{\pi}3\right]\times \left(\left[0,\frac{\pi}2\right]\setminus B\right)\right)$, and conclude since $\displaystyle \left[0,\frac{\pi}2\right]\setminus B$ is countable.