The graph of every real function has inner measure zero

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).