Find an example for a bounded function $f$ where the graph of $f$ marked by ${E=(x,f(x))}$ is not of volume 0.
I tried to think about some variation of dirichlet function but the I saw that you could treat the graph as two seperate graphs of continuous functions and bound them..
Here's my counterexample. You might need to flesh it out a bit.
Let $I = [0, 1] \cap (\mathbb{R} \setminus \mathbb{Q})$. Clearly, $|I| = \aleph$ and $\partial{I} = [0,1]$.
Therefore, there is a disjoint collection of dense, countable subsets of $I$, $\{I_n: n\in\mathbb{N} \}$. To see why, take a dense, countable subset $I_1$ of $I$, and because $|I\setminus I_1|=\aleph$ and $\partial (I\setminus I_1)=[0,1]$, we can construct the desirable collection by induction.
Now, since $\mathbb{Q}\cap [0,1]$ is countable there is a bijective map $g: \mathbb{N} \to \mathbb{Q}\cap [0,1]$. Then let $f:\bigcup_{n\in\mathbb{N}} I_n \to \mathbb{R}$ be defined $f(x)=g(n)$ for all $x\in I_n$.
And then clearly $\partial{E} = [0,1]^2$ for $E=\{(x,f(x))\in\mathbb{R}^2\}$. And therefore, $E$ is not even (Jordan) measurable, let alone has volume $0$.
Hope that helps. Good luck on Tuesday.