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Thursday, May 30, 2013

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

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