First of all, sorry for asking again a question.
For all functions, $f:[a,b] \to \mathbb{R}_{+} $ define
$S(f)=\{(x,y)\in \mathbb{R}^{2}: 0\leq y \leq f(x)\}$.
Show if $f$ is measurable, then $S(f) \in \mathbb{B} (\mathbb {R}^{2})$. Compute that two-dimensional Lebesgue measure of $S(f)$.
For the first part, I thought maybe if I define a function $g$ such that $g(x,y)=f(x)-x$ and if I can show this is continuous , it may work.. will it ?
For the second part, can I approach with integrable simple functions ?..
Since I don't have good knowledge of this subject , I don't know what to do anymore. Maybe it will be good to push me a little just by giving hints not the answer. Thanks in advance.
For the first part, let $\phi(x,y) = f(x)-y$. Then $\phi$ is measurable and $S(f) = (\mathbb{R} \times [0,\infty) ) \cap \phi^{-1} [0,\infty)$ and both $\mathbb{R} \times [0,\infty)$ and $\phi^{-1} [0,\infty)$ are measurable, hence $S(f)$ is measurable.
For the second, use Fubini-Tonelli (I used $m$ to denote the product measure), and note that $1_{S(f)}(x,y) = 1_{[0,f(x)]}(y)$.
Then $\int 1_{S(f)} = \int 1_{S(f)}(x,y) dm = \int \left( \int 1_{S(f)}(x,y) dy \right) dx = \int \left( \int 1_{[0,f(x)]}(y) dy \right) dx = \int f(x) dx = \int f.$