The graph of every real function has inner measure zero I'm trying to understand the answer to this question:

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a (general) function. Is there an $N\subset \mathbb{R}^2$ with $\lambda^2(N)=0$, such that $\{(x,f(x)):x\in \mathbb{R}\}$ $\subset N$ ?

First line of the answer, found there: "No function can have a graph with positive measure or even positive inner measure, since every function graph has uncountably many disjoint vertical translations, which cover the plane. "

I don't understand why the fact that "every function graph has uncountably many disjoint vertical translations, which cover the plane" (I agree with that) implies the desired result.

Thank you for your help.
 A: If a function graph $G$ of positive inner measure existed, then choosing $K\subset G$ measurable of positive measure and considering the translates of $K$, you would get uncountably many disjoint measurable sets of positive measure.  This is impossible in any $\sigma$-finite measure space.
Indeed, let $X=\bigcup X_n$ be a $\sigma$-finite measure space with $\mu(X_n)<\infty$ and let $\mathcal{C}$ be an uncountable collection of disjoint measurable subsets of $X$ of positive measure.  Then for each $A\in\mathcal{C}$, there exist $m,n\in\mathbb{N}$ such that $\mu(A\cap X_n)>1/m$.  Thus there must exist some pair $(m,n)\in\mathbb{N}^2$ such that $\mu(A\cap X_n)>1/m$ for uncountably many different elements $A\in \mathcal{C}$.  But since the sets $A\cap X_n$ are all disjoint, this would imply $\mu(X_n)=\infty$ (say by choosing a countably infinite collection of such $A$ and using countable additivity).
A: Suppose the graph had positive measure. Then we can cover the plane with uncountably many disjoint translates of this positive-measure set. Then this question completes the proof.
