Show that $\operatorname{Vol}_{3}(\{0\leq x,y\leq1,\,z=2x+y\})=0$

Question

I want to show that $\operatorname{Vol}_{3}=0$, where $$D:=\{(x,y,z)\in\mathbb{R}^{3}\,:\,0\leq x,y\leq1,\,z=2x+y\}$$ by covering $D$ (for each $\epsilon>0$) with a finite number of rectangles $R_{i}$ such that $\sum_{i}\operatorname{Vol}_{3}(R_{i})<\epsilon.$

What sort of choice of rectangles will be suitable here?

I've done a few examples in $\mathbb{R}^{2}$, such as the line $y=x$ from $(0,0)$ to $(1,1)$, but I am having trouble extending this mindset in $\mathbb{R}^{3}$.

Any help is welcome!

Answer 1

HINT: Divide $[0,1]\times [0,1]$ into $n\times n$ squares. Over the $i$th square, you want a box $R_i$ of height $<\epsilon$ that encloses the portion of $D$ over that square. So you want to choose $n$ large enough so that the maximum of $2x+y$ minus the minimum of $2x+y$ on each square will be $<\epsilon$. Can you figure this out? (Spoiler: You will need $3/n<\epsilon$.)

Answer 2

Hint.

The set $D$ is the graph of a Riemann integrable function $f$ on a square $[0,1]^2$.

If you consider the sets of "3d rectangles" that give you the lower sum and the upper sum, the set difference between the "upper rectangles" and the "lower rectangles" contains the set $D$. Because of integrability, the difference between the lower sum and the upper sum can be arbitrarily small.

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Lower dimension analogue: