Show that the following Sets have Lebesgue_Measure Zero I want to show that the following two sets have Lebesgue-Measure Zero.
1.) The graph($f$) in $\mathbb R^2$ if $f$ is given as $f:\left[ 0,1 \right] \rightarrow \mathbb R$.
2.)$\left\{ (x,y,z):\quad x+y+z=0 \right\} \quad in\ \mathbb R^3$. 
What's need to be done is to find open Cubes so that the Sum over the Volume of those open Cubes is smaller than an abitrary Epsilon.
And I'm havin struggels to construct those open Cubes.
 A: $\newcommand\restrict[1]{\raise{-.5ex}{\big|}_{#1}}\newcommand\gra[1]{\text{graph}\kern-0.12pc\left(#1\right)}%\newcommand\restr[2]{{ \left.\kern0pc #1 \vphantom{\big|} \right|_{#2} }} \restr{f}{A} f\restrict{[0,\epsilon]}$
Any continuous function in a compact set is uniformly continuous (show it!).
This means that
$$(\forall\epsilon>0)\,(\exists\delta>0)\,(\forall x,y\in[0,1] \text{ with $|x-y|\leq\delta$})\, |f(x)-f(y)|\leq\epsilon.$$
Try to picture this.
Let $\gra{f}$ denote the graph of $f$.
Fix some $\epsilon>0$, and let $\delta$ be an associated $\delta$  obtained from uniform continuity.
We can cover the interval $[0,1]$ in $n_\delta=\left\lceil\frac1{2\delta}\right\rceil$ intervals of length $2\delta$.
Notice that $n_\delta<\frac1{2\delta}+1$.
Now, let $I$ be any such interval and $x_I\in I$ be its midpoint.
For all $y\in I$ we have $|x_I-y|<\delta$ so uniform continuity guarantees that
$$f(y)\in[f(x_I)-\epsilon,f(x_I)+\epsilon].$$
This means that $\gra{f\restrict{I}}\subset
I\times[f(x_I)-\epsilon,f(x_I)+\epsilon]$, and hence
$$\mu\left(\gra{f\restrict{I}}\right)\leq 2\delta\times2\epsilon=4\epsilon\delta.$$
Since we can cover $[0,1]$ with $n_\delta$ such intervals, we conclude that
\begin{align}
\mu\left(\gra{f}\right)
&\leq 4\epsilon\delta\cdot n_\delta\\
&\leq 4\epsilon\delta\left(\frac1{2\delta}+1\right)\\
&=2\epsilon+4\epsilon\delta\\
&\leq 6\epsilon
\end{align}
In the last step, we have used that $\delta\leq1$.
The conclusion follows.
