1.17

The goal of this exercise is to show that if we choose $c_n$'s as the hint prescribes, the set: $$E = \Big\lbrace x \in [0,1] \hspace{0.25cm} \bigg| \hspace{0.25cm} \lim_{n \to \infty} \Big| \frac{f_n(x)}{c_n} \Big| > 0 \Big\rbrace $$ has measure zero. Define: $$ S_c = \Big\lbrace x \in [0,1] \hspace{0.25cm} \bigg| \hspace{0.25cm} |f_n(x)| > \frac{c}{n} \Big\rbrace$$ Notice that $m(S_1) \leq 1$. Also, notice $S_{c+1} \subset S_{c} \hspace{0.25cm} \forall c$, and $\bigcap_{c=1}^{\infty}S_c = \emptyset$. Thus, by Corollary 3.3, $\lim_{n \to \infty} m(S_n) = 0$.

With that out of the way, notice that for the sets: $$E_n = \Big\lbrace x \in [0,1] \hspace{0.25cm} \bigg| \hspace{0.25cm} \Big| \frac{f_n(x)}{c_n} \Big| > \frac{1}{n} \Big\rbrace $$ ...we can choose $c_n$ large enough such that for every $n$, $m(E_n) \leq \frac{1}{2^n}$. Certainly, it follows that ($\dagger$): $$ \sum_{n=1}^{\infty} m(E_n) \leq \sum_{n=1}^{\infty} \frac{1}{2^n} = 1 < \infty $$ Now, observe that $E = \sup{ \lbrace E_n \rbrace }$, i.e. (recalling problem 1.16), $E = \bigcap_{k=1}^{\infty} \bigcup_{n \geq k}E_n$. Thus, applying the Borel-Cantelli Lemma, we have that $$ (\dagger) \Rightarrow m(E) = 0$$