The region under a non-negative measurable function

Question

Let $f$ be a non-negative measurable function on $E\subseteq\mathbb{R}^d$ and let $$R(f)=\{(x,y)\in\mathbb{R}^{d+1}:x\in E,0\leq y\leq f(x)\}.$$ Then it's a well-known result in Real Analysis that $R(f)$ is measurable. If we define $$\bar{f}(x)=\begin{cases} f(x)&x\in E\\ 0&x\notin E, \end{cases}$$ and $m$ denotes the Lebesgue measure in $\mathbb{R}^{d+1}$, then is it possible to prove that $m(R(f))=m(R(\bar{f}))$? Thank you.

Answer 1

By Fubini\Tonelli's Theorem $m(R(f))=\int_E \int_0^{f(x)} dydx=\int_E f(x)dx$. Since $\int_{\mathbb R^{d}} \overline {f} (x)dx=\int_E f(x)dx$ we get $m(R(f))=m(R(\overline {f}))$.