I am proving the following theorem:
For a subset $E$ of $\mathbb{R}^n$, if $E$ is measurable, then $E_a$ is measurable and $|E_a| = a|E|$.
where $E_a := \{(\mathbf{x}, y):\mathbf{x}\in E, 0\le y\le a\}$ for finite $a$,
$E_\infty := \{(\mathbf{x}, y):\mathbf{x}\in E, 0\le y\lt+\infty\}$,
$|E|_e$ is outer measure of $E$,
and $|E|$ is measure of $E$.
My textbook said:
If $|E|=0$ or if $E$ is an interval which is either closed, partly open, or open, the result is clear.
Actually, It seems that if $E$ has measure $0$, then $E_a$ from intuitive. However I want to know why it follows. Can someone let me know the reason why "if $|E|=0$ or if $E$ is an interval which is either closed, partly open, or open, then $E_a$ is measurable and $|E_a|=a|E|$."
$E_a$ is just $E \times [0,a] \subset \mathbb{R}^{n+1}.$ And $|E \times [0,a]| = |E| \times |[0,a]|$ by the definition of product measure.