MHB Riemann integral is zero for certain sets

[i_a_n]

The question is:Let $\pi=\left \{ x\in\mathbb{R}^n\;|\;x=(x_1,...,x_{n-1}, 0) \right \}$. Prove that if $E\subset\pi$ is a closed Jordan domain, and $f:E\rightarrow\mathbb{R}$ is Riemann integrable, then $\int_{E}f(x)dV=0$.(How to relate the condition it's Riemann integrable to the value is $0$? The textbook I use define $f$ is integrable on $E$ iff $\;\;\;\;(L)\int_{E}fdV=(U)\int_{E}fdV$)

 

[Jhoneine]

Answer:Since $E \subset \pi$, we know that $E$ is a set of points with coordinates of the form $(x_1, x_2, ... , x_{n-1}, 0)$. This implies that the last coordinate of any point in $E$ is zero. Therefore, the volume element for this set of points is $dV=dx_1\,dx_2\,...\,dx_{n-1}$.Now, since $f$ is Riemann integrable, we know that $(L)\int_{E}fdV=(U)\int_{E}fdV$ and thus the integral of $f$ over $E$ is equal to the lower bound and upper bound of the integral. Since the last coordinate of any point in $E$ is zero, the integral of $f$ over $E$ can be written as $$\int_{E}f(x)dV=\int_{E}f(x_1,x_2,...,x_{n-1})dx_1\,dx_2\,...\,dx_{n-1}.$$Since the last coordinate of any point in $E$ is zero, we can set the integrand to be zero and the integral becomes$$\int_{E}f(x)dV=0.$$Therefore, we have shown that if $E\subset\pi$ is a closed Jordan domain, and $f:E\rightarrow\mathbb{R}$ is Riemann integrable, then $\int_{E}f(x)dV=0$.