User:Kcbetancourt/AnalysisHW5 1

4.1 a.) ''Show that if $$ f(x) = \begin{cases} 0, & x \notin \mathbb{Q}            \\ 1, & x \in \mathbb{Q}  \end{cases} \ $$

then $$ R\overline{\int_a^{b}} f(x) dx = b-a $$ and $$ R\underline{\int_a^b} f(x) dx = 0 $$.''

Let $$ a = \xi_0 < \xi_1 < \cdots < \xi_n = b $$ be a subdivision of $$ [a,b]$$. Then for each subdivision, $$ S = \sum_{i=1}^{n}(\xi_1 - \xi_{i-1})M_i$$ and $$ s = \sum_{i=1}^{n}(\xi_1 - \xi_{i-1})m_i$$ where $$ M_i = \sup_{\xi_{i-1}<x\le \xi_i} f(x)$$ and $$ m_i = \inf_{\xi_{i-1}<x\le \xi_i} f(x)$$. We know that $$ \sup f(x) = 1 $$ and $$ \inf f(x) = 0$$, $$ \forall i $$. Therefore $$ S = b-a$$ and $$ s = 0 $$ for any subdivision of $$ [a,b]$$. Then, by definition, $$ R\overline{\int_a^{b}} f(x) dx = \inf S $$ which is equal to $$ b-a $$ and $$ R\underline{\int_a^{b}} f(x) dx = \sup s $$ which is equal to $$ 0 $$. Thus, $$ R\overline{\int_a^{b}} f(x) dx = b-a $$ and $$ R\underline{\int_a^b} f(x) dx = 0 $$.

b.) Construct a sequence $$ \left\{{f_n}\right\} $$ of nonnegative, Riemann integrable functions such that $$ f_n $$ increases monotonically to $$ f $$. What does this imply about changing the order of integration and the limiting process?

Let $$ \left\{{q_n}\right\} $$ be a list of all rational numbers in the interval $$ [a,b] $$. Then define $$ \left\{{f_n}\right\} = \chi_{q_1, \ldots ,q_n}$$. Then $$ \left\{{f_n}\right\} $$ is a sequence of nonnegative Riemann integrable functions increasing monotonically to $$ f $$. We can take the limit of the integral, but we may not be able to integrate the limit. Therefore, in general, we cannot change to order of integration and the limiting process.

4.3 ''Let $$ f $$ be a nonnegative measurable function. Show that $$ \int f= 0 $$ implies that $$ f= 0 $$ a.e.''

Define $$ g_n (x) = \begin{cases} \frac{1}{n}, & f(x)\ge \frac{1}{n} \\ 0, & otherwise \end{cases} \ $$

Then, $$ 0\le g_n(x)\le f(x), \forall n,x $$. Then, $$ \forall n $$ we have $$ 0\le \int g_n \le \int f $$. Since $$ \int f = 0 $$, $$ \int g_n = 0$$, $$ \forall n $$. Let $$ E_n $$ be the set of all $$ x $$ for which $$ f(x)\ge \frac{1}{n} $$. Then $$ \int g_n = \frac{1}{n} = mE_n $$. But, $$ \int g_n = 0 $$. Therefore, $$ mE_n = 0 $$. Now let $$ E = \bigcup_{n} E_n $$. Then, since measure is countably additive, $$ mE = 0 $$. Therefore, $$ E $$ is the set of all $$ x $$ such that $$ f(x) $$ is non-zero, which has measure zero. Thus, $$ f = 0 $$ almost everywhere.

4.8 Prove the following generalization of Fatou's Lemma: If $$ \left\{{f_n}\right\} $$ is a sequence of nonnegative functions, then $$ \int \lim \inf f_n \le \lim \inf \int f_n $$.

By definition of lim inf, $$ \lim \inf f_n = \sup_n \inf_{k\ge n} f_k $$. Let $$ g_n = \inf_{k\ge n} f_k $$. Then, $$ g_n\le f_n $$, which implies that $$ \int g_n\le \int f_n $$, and $$ \lim g_n = \lim \inf f_n $$. Also, each $$ g_n $$ is a nonnegative measurable function. Then, $$ \int \lim \inf f_n = \int \lim g_n =\lim \int g_n \le \lim \inf \int g_n \le \lim \inf \int f_n $$. Therefore, $$ \int \lim \inf f_n \le \lim \inf \int f_n $$.

4.15

a.) Let $$ f $$ be integrable over $$ E $$. Then, given $$ \varepsilon > 0 $$, there is a simple function $$ \phi $$ such that $$ \int_E \left|{f-\phi }\right| < \varepsilon $$. [Apply Problem 4 to the positive and negative parts of $$ f $$.]

Problem 4 tells us that there exists a simple function $$ \phi_1 \le f $$ such that $$ \int_E f - \frac{\varepsilon}{2} < \int_E \phi_1 $$ for positive $$ f $$, call it $$ f^+ $$. And there also exists a simple function $$ \phi_2 \le f $$ such that $$ \int_E f - \frac{\varepsilon}{2} < \int_E \phi_2 $$ for negative $$ f $$, call it $$ f^- $$. Then $$ \int f = \int (f^+ - f^-) $$. Now define a simple function $$ \phi = \phi_1 - \phi_2 $$. Then $$ \int_E \left|{f-\phi }\right| = \int_E \left|{f^+ -\phi_1 - f^- + \phi_2}\right| \le \int_E \left|{f^+ -\phi_1 }\right| + \int_E \left|{-(f^- -\phi_2) }\right| = \int_E f^+ - \phi_1 + \int_E f^- - \phi_2 = \int_E f^+ - \int_E \phi_1 + \int_E f^- - \int_E \phi_2 < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon $$. Therefore, $$ \int_E \left|{f-\phi }\right| < \varepsilon $$.

b.) Under the same hypothesis there is a step function $$ \psi $$ such that $$ \int_E \left|{f-\psi }\right| < \varepsilon $$. [Combine part (a) with Proposition 3.22.]

Define $$ f_n = f\chi_{[-n,n]} $$. Then we have $$ \left|{f_n}\right| \le \left|{f}\right| $$ and $$ f_n\to f $$. By the Lebesgue Dominated Convergence Theorem, $$ \int_E f = \lim_{n\to \infty} \int_E f\chi_{[-n,n]} $$. This implies that $$ \int_E \left|{f - f\chi_{[-n,n]} }\right| \to 0 $$, and thus $$ \int_{E\bigcap [-n,n]^c} \left|{f}\right| \to 0 $$. Then $$ \exists N : \int_{E\bigcap [-N,N]^c} \left|{f}\right| < \frac{\varepsilon}{3} $$. From part (a), we know that there is a simple function $$ \phi $$ such that $$ \int_E \left|{f-\phi }\right| < \frac{\varepsilon}{3} $$. And by Prop 3.22, there exists a step function $$ \psi $$ on the interval $$ [-N,N] $$ such that $$ \left|{\phi - \psi}\right| < \frac{\varepsilon}{12NM} $$ except on a set of measure less than $$ \frac{\varepsilon}{12NM} $$, where $$ M \ge \max (\left|{\phi}\right|,\left|{\psi}\right|)+1 $$. Let $$ \psi = 0 $$ outside the interval $$ [-N,N] $$. Then we have, $$ \int_{-N}^N \left|{\phi - \psi}\right| < \frac{\varepsilon}{3} $$. There for we can write $$ \int_E \left|{f-\psi }\right| \le \int_E \left|{f-\phi }\right| + \int_{-N}^N \left|{\phi - \psi}\right| + \int_{E\bigcap [-N,N]^c} \left|{f}\right| < \frac{\varepsilon}{3} +\frac{\varepsilon}{3} +\frac{\varepsilon}{3} = \varepsilon $$. Thus, $$ \int_E \left|{f-\psi }\right| < \varepsilon $$.

c.) Under the same hypothesis there is a continuous function $$ g $$ vanishing outside a finite interval such that $$ \int_E \left|{f-g}\right| < \varepsilon $$.

From part (b), we know that there exists a step function $$ \psi $$ such that $$ \int_E \left|{f-\psi }\right| < \varepsilon $$. Let $$ \psi $$ be defined on the finite interval $$ [a,b] $$ such that $$ \psi = 0 $$ anywhere outside the interval. Define $$g $$ to be the continuous function formed by linearising $$ \psi $$. Then $$g $$ vanishes outside the finite interval such that $$ g = \psi $$. Then $$ \int_E \left|{f-g}\right| \le \int_E \left|{f-\psi}\right| + \int_E \left|{\psi -g}\right| < \varepsilon + 0 $$. Therefore, $$ \int_E \left|{f-g}\right| < \varepsilon $$.

16. Establish the Riemann-Lebesgue Theorem: If $$ f $$ is an integrable function on $$ (-\infty, \infty ) $$, then $$ \lim_{n\to \infty} \int_{-\infty}^\infty f(x) \cos nx dx = 0 $$. ''[Hint: The theorem is easy if $$ f $$ is a step function. Use Problem 15.]''

Let $$ f $$ be an integrable function on $$ (-\infty, \infty ) $$ and let $$ \varepsilon > 0 $$. Define a step function $$ \psi $$ such that $$ \int_E \left|{f-\psi }\right| < \frac{\varepsilon}{2} $$. Then $$ \left|{ \int f(x) \cos nx dx}\right| \le \int \left|{f(x) \cos nx }\right|dx \le \int \left|{(f(x) - \psi (x)) \cos nx }\right|dx + \int \left|{\psi (x) \cos nx }\right|dx < \frac{\varepsilon }{2} + \int \left|{\psi (x) \cos nx }\right|dx $$. Integrating, as $$ n\to \infty $$ we get $$ \int \left|{\psi (x) \cos nx }\right|dx \to 0 $$. Then, $$ \exists N : \forall n\ge N, \int \left|{\psi (x) \cos nx }\right|dx < \frac{\varepsilon}{2} $$. Therefore, $$ \left|{ \int f(x) \cos nx dx}\right| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon $$ and $$ \lim_{n\to \infty} \int f(x) \cos nx dx = 0 $$.