1.37

The exercise defines: $$\Gamma = \lbrace (x,y) \in \mathbb{R}^2 \hspace{0.25cm} \big| \hspace{0.25cm} y = f(x), x \in \mathbb{R} \rbrace$$ For any $k \in \mathbb{Z}$, let $\hspace{0.25cm} I_k = [k,k+1]$.

Since $f$ is continuous on $\mathbb{R}$, $f$ is actually uniformly continuous an any compact interval. This means that on any $I_k$, $\hspace{0.25cm} \forall \epsilon > 0, \hspace{0.25cm} \exists \delta > 0$ such that $|x - y| < \delta \Rightarrow \hspace{0.25cm} |f(x) - f(y)| < \epsilon$.

For any $\epsilon > 0$, choose $m$ large enough such that $$|x - y| \leq \frac{1}{2^{m + |k|+2}} \hspace{0.25cm} \Rightarrow \hspace{0.25cm} |f(x) - f(y)| < \epsilon$$

Next, let $\lbrace I_k^j \rbrace$ be a collection of compact intervals of length $\frac{1}{2^{m + |k|+2}}$ such that we can write: $$I_k = \bigcup_{j=1}^{2^m} I_k^j$$ Next, notice, if: $$\Gamma_k = \lbrace (x,y) \in \mathbb{R}^2 \hspace{0.25cm} \big| \hspace{0.25cm} y = f(x), x \in I_k \rbrace$$ Then: $$\Gamma_k \subset \bigcup_{j=1}^{2^m} \big(I_k^j \times f(I_k^j)\big)$$ And: $$m(\Gamma_k) \leq m\Big(\bigcup_{j=1}^{2^m} \big(I_k^j \times f(I_k^j)\big)\Big) \leq \sum_{j=1}^{2^m} m\big(I_k^j \times f(I_k^j)\big) \leq \ldots$$ $$\ldots \leq \sum_{j=1}^{2^m} \frac{\epsilon}{2^{m + |k|+2}} = \frac{2^m \epsilon}{2^{m + |k| + 2}} = \frac{\epsilon}{2^{|k| + 2}} $$ Lastly, notice that $\Gamma = \bigcup_{k=-\infty}^{\infty} \Gamma_k$. I.e. $$m(\Gamma) \leq \sum_{k=-\infty}^{\infty} m(\Gamma_k) \leq \sum_{k=-\infty}^{\infty} \frac{\epsilon}{2^{|k| + 2}} \leq \ldots$$ $$\ldots \leq 2\sum_{k=0}^{\infty} \frac{\epsilon}{2^{|k| + 2}} = 2\frac{\epsilon}{2} = \epsilon$$ Thus, since $\epsilon > 0$, arbitrarily, we have that $m(\Gamma) = 0$.