I have no idea about how the Volume of a set works, It would be great if anyone can help me. This is the definition of Vol(A) I'm working with enter image description here

Let $f: \mathbb{R} \rightarrow [0,\infty)$ a density function and Let $A=\left\{(x,y)\in \mathbb{R} \times [0,\infty) ; 0<y<f(x)\right\}$ show that Vol(A)=1

  • $\begingroup$ Unless I'm mistaken (disclaimer: I have no familiarity with measure theory beyond undergraduate-level probability), the "volume" of a two-dimensional set is the area of the set. Volume is sometimes used interchangeably with measure. $\endgroup$ – user170231 Feb 13 at 22:11
  • $\begingroup$ @user170231 would you please give me a hint? $\endgroup$ – Eduardo Cuéllar Feb 13 at 22:14
  • $\begingroup$ Use the fact that the indicator function $1_A(x, y)$ is $1$ if and only if $0 < y < f(x)$ to rewrite the double integral into the single integral $\int_{\mathbb R} f(x) dx$, which is equal to $1$ since $f$ is a probability density function. (I assume your $F$ is meant to be $f$, otherwise we need more information) $\endgroup$ – Rushy Feb 13 at 22:52
  • $\begingroup$ @Rushy thanks a lot! But, why is it valid to change the double integral to just the integral of the density function? I had that idea but I wasn't sure why. $\endgroup$ – Eduardo Cuéllar Feb 13 at 23:18
  • $\begingroup$ You can probably use Tonelli's theorem together with $1_A$ being a non-negative measurable function, although there are probably easier ways (my analysis is a bit rusty) $\endgroup$ – Rushy Feb 13 at 23:35

\begin{align*} {\rm Vol}(A)&=\int_{{\mathbb R} \times [0,\infty)}{\bf 1}_{A}(x,y){\rm Vol}({\rm d}(xy)) \\ &=\int_{{\mathbb R}}\int_{0}^{\infty}{\bf 1}_{A}(x,y){\rm Vol}({\rm d}y){\rm Vol}({\rm d}x) \\ &=\int_{{\mathbb R}}\int_{0}^{\infty}{\bf 1}_{(0,f(x))}(y){\rm Vol}({\rm d}y){\rm Vol}({\rm d}x) \\ &=\int_{{\mathbb R}}{\rm Vol}((0,f(x))){\rm Vol}({\rm d}x) \\ &=\int_{{\mathbb R}}f(x){\rm Vol}({\rm d}x) \\ &=1. \end{align*}

\begin{align*} {\bf 1}_{A}(x,y)&= \left\{ \begin{array}{ll} 1, & (x,y) \in A \\ 0, & \text{otherwise} \end{array} \right. \\ &= \left\{ \begin{array}{ll} 1, & 0<y<f(x) \\ 0, & \text{otherwise} \end{array} \right. \\ &= \left\{ \begin{array}{ll} 1, & y \in (0,f(x)) \\ 0, & \text{otherwise} \end{array} \right. \\ &={\bf 1}_{(0,f(x))}(y). \end{align*}

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    $\begingroup$ Thanks a lot! But I have one question... What happens between the second and third line? (When we change $1_{A}(x,y)$ to $1_{A}(0.f(x))(y)$). I'm just started with integrals in $\mathbb{R}^{n}$, I will really appreciate it! $\endgroup$ – Eduardo Cuéllar Feb 17 at 17:22
  • $\begingroup$ The proof was added. $\endgroup$ – 720773 Feb 18 at 2:51

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