2.7

First, assume $f$ is almost-everywhere finite. Next, partition $\mathbb{R}^d$ into almost-disjoint closed unit cubes $\lbrace Q_k \rbrace_{k=1}^\infty$. Similarly to how the exercise defines $\Gamma$, define: $$\Gamma_k = \lbrace (x,y) \in Q_k \times \mathbb{R} \hspace{0.25cm} \big| \hspace{0.25cm} y = f(x) \rbrace$$ Next, define the sets: $$F_{k,n}^i = \lbrace x \in Q_k \hspace{0.25cm} \big| \hspace{0.25cm} \frac{i}{2^n} \leq f(x) < \frac{i+1}{2^n} \rbrace$$ Then, define: $$E_{k,n}^i = F_{k,n}^i \times \Big[\frac{i}{2^n}, \frac{i+1}{2^n}\Big)$$ And, finally: $$E_{k,n} = \bigcup_{i = -\infty}^{\infty} E_{k,n}^i $$ All of the above sets clearly inherit their measurability from $f$. Notice the following: $$\Gamma_k \subset E_{k,n} \hspace{0.25cm} \forall n \in \mathbb{N}$$ and ($\dagger$), $$E_{k,n+1} \subset E_{k,n} \hspace{0.25cm} \forall n \in \mathbb{N}$$ Now, observe that $$m^{d+1}(E_{k,n})\hspace{0.25cm} \leq \hspace{0.25cm}\sum_{i=-\infty}^\infty m^{d+1}(E_{k,n}^i) \leq \ldots$$ $$\ldots \leq \sum_{i=-\infty}^\infty m^{d}(F_{k,n}^i) \cdot m\Big(\big[\frac{i}{2^n}, \frac{i+1}{2^n}\big)\Big) = \ldots$$ $$\ldots = \frac{1}{2^{n-1}}\sum_{i=-\infty}^\infty m^{d}(F_{k,n}^i) \hspace{0.25cm} \leq \hspace{0.25cm} \frac{1}{2^{n-1}} \cdot m^d(Q_k) = \frac{1}{2^{n-1}}$$ Observe, by $\dagger$, that since the sets $E_{k,n}$ are collapsing, with finite measure, that $\Gamma_k$ must be measurable since: $$m^*(\Gamma_k) \leq \lim_{n \to \infty} m(E_{k,n}) = \lim_{n \to \infty} \frac{1}{2^{n-1}} = 0$$ Thus,each $\Gamma_k$ is measurable, which implies $\Gamma$ is measurable, since $\mathcal{M}$ is closed under countable unions. Lastly, we just need to observe that $$m^{d+1}(\Gamma) \leq \sum_{k = 1}^{\infty} m^{d+1}(\Gamma_k) = 0$$ ...which is what we set out to show.