Measure of $M:= {(x,y) \in [0,1]^2 : y^2 \leq x } $ How can I find (jordan-) measure of 
$M:= \{(x,y) \in [0,1]^2 : y^2 \leq x \} $
?
what would be different to the Lebesgue-measure?
 A: Let $S$ be a half-open rectangle in $\mathbb R^2$. Jordan measure on an arbitrary bounded set $A$ in the plane is defined by considering the following set functions: 
$\tag 1c^*(A)=\inf \sum \text{vol }S$ and 
$\tag2 c_*(A)=\sup \sum \text{vol }S$
where the sums are all $finite$ and taken over unions of rectangles that cover $A$ in the first case, and are contained in $A$ in the second case.
The Jordan measure of $A$ is defined to be $c(A)=c_*(A)=c^*(A)$. whenever $(1)$ and $(2)$ agree.
It is easy to see that if $f:[0,1]\to \mathbb R$ is Riemann integrable, then $A=\left \{ 0\le x\le 1;\ \le 0\le y\le f(x) \right \}$ is Jordan measurable, because in this case 
$\tag 3 c^*(A)=\overline \int_{[0,1]} f dx$ 
and
$\tag 4 c_*(A)=\underline \int_{[0,1]} f dx$
In fact, the Jordan measure on $\mathbb R^n$ is sometimes defined as $\mathscr R\int_{\mathbb R^{n}} \chi_A(\vec x)d\vec x,\ $ whenever this integral exists. 
Now, since $f(x)=\sqrt x$ is Riemann integrable on $[0,1]$, it is Lebesgue integrable there, and so by our previous remarks, $M$ is Jordan and Lebesgue measurable, these are equal and the value can be calculated by evaluating $\int^1_0\sqrt xdx=\frac{1}{2}.$
A: According to https://en.wikipedia.org/wiki/Jordan_measure the onnly difference is that we should consider the interior of the set instead of whole of it. In this case:$$int(S)=\{(x,y)|0<x,y<1,y^2<x\}$$ from which turns out that$$m_J(S)={2\over 3}$$
