User:J D Bowen/Math710 HW5

4.1 a.) We aim to 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 $.

Observe that for any partition $a=x_0<x_1<\dots<x_n=b \ $ of $[a,b], \forall n\in\left\{{1,n}\right\}, \ \exists q_n,r_n \in [x_{n-1},x_n] \ $ such that $q_n\in \mathbb{Q}, r_n\not\in\mathbb{Q} \ $.

Then $\forall i, \ M_i=1, \ m_i =0 \ $, and so $S=b-a, \ s=0 \ $. Since this is for any partition,

$R\overline{\int_a^b} f(x)dx = \text{inf}(S)=b-a \ $,

$R\underline{\int_a^b} f(x)dx = \text{sup}(s)=0 \ $.

b.) We aim to construct a sequence $ \left\{{f_n}\right\} $ of nonnegative, Riemann integrable functions such that $ f_n $ increases monotonically to $ f $, and infer from this an implication regarding changing the order of integration and the limiting process. So let $\left\{{y_k}\right\}_{k=1}^\infty \ $ be a list of rational numbers in $[a,b] \ $, and define

$f_n(x)=\begin{cases} 0, & x \notin \left\{{y_k}\right\}_{k=1}^n \\ 1, & x \in \left\{{y_k}\right\}_{k=1}^n \end{cases} \ $

Then $\forall x, \ m\geq n \implies f_m(x)\geq f_n(x), \ f_n\to f \ $.

Observe that for each $n, \ f_n(x)=0 \ $ except at a finite number of points. We need to put these points in order to get the partition we want, so start by defining $z_0 =a, \ z_{n+1}=b, \ $ and then define $z_1=\text{min}(\left\{{y_k}\right\}_{k=1}^n), \ z_2 =\text{min}(\left\{{y_k}\right\}_{k=1}^n\backslash\left\{{z_1}\right\}), z_3=\text{min}(\left\{{y_k}\right\}_{k=1}^n\backslash\left\{{z_1,z_2}\right\}) \ $, and in general, $z_k=\text{min}(\left\{{y_l}\right\}_{l=1}^n\backslash\left\{{z_j}\right\}_{j=1}^{k-1}) \ $.

Now let $\epsilon>0 \ $ and set $x_0 = a, \ x_{2n+2}=b \ $ and set $x_{2k-1}=z_k-\tfrac{\epsilon}{2n}, \ x_{2k}=z_k+\tfrac{\epsilon}{2n} \ $ for $1\leq k \leq n \ $. Then, supposing $\epsilon<(1/2)\text{min}|x_n-x_{n-1}| \ $ (a reasonable assumption, since we will need epsilon to be small, not large) we have a partition $a=x_0<x_1<\dots<x_{2n+2}=b \ $. Call this the epsilon-partition.

Note that for this partition,

$S=\sum_{i=1}^{2n+2} M_i (x_i-x_{i-1}) = 0+1\cdot \frac{\epsilon}{n}+0+1\cdot\frac{\epsilon}{n}+\dots = n\frac{\epsilon}{n}=\epsilon \ $,

and

$s=\sum_{i=1}^{2n+2} M_i (x_i-x_{i-1})= 0 \ $.

Therefore, $R\overline{\int_a^b} f(x)dx = \text{inf}(S)=\text{inf}(\epsilon)=0 \ $ and

$R\underline{\int_a^b} f(x)dx = \text{sup}(s)=0 \ $.

Hence the integral exists. Since we have constructed a sequence of integrable functions converging to a non-integrable one, we have shown limits cannot pass in or out of Riemann integrals, in general.

4.3 Given a nonnegative measurable function $ f $, we aim to show that $ \int f= 0 \implies f= 0  \ $ almost everywhere.

Define $S=\left\{{x:f(x)>0 }\right\} \ $. What we want to do is say the integral is greater than the minimum value of f on this set, times its measure, and then observe that the integral is zero, implying the measure is zero. But since we don't know what the minimum value is on this set, or if such a value exists, define

$S_n=\left\{{x:f(x)>\tfrac{1}{n} }\right\} \ $

and observe that $S=\cup S_n \ $. Now, since $\int f \geq \tfrac{1}{n}mS_n \implies mS_n=0 \ $, we can find $mS \ $ as

$mS = m\left({ \bigcup_{n=1}^\infty S_n }\right) = \sum_{k=1}^\infty mS_n = \Sigma 0 = 0 \ $.

4.5 Let $ f \ $ be a nonnegative integrable function. We aim to show that the function $ F \ $ defined by $ F(x) = \int_{-\infty}^{x} f \ $ is continuous by using the Monotone Convergence Theorem. Recall that the MCT states that given an increasing sequence of nonnegative measurable functions $ \left\{{f_n}\right\} \ $ with $ f = \lim f_n \ $ almost everywhere, we have $ \int f = \lim\int f_n \ $.

Define $f_n = f|_{(-\infty,x-\tfrac{1}{n}]}, f^*_n=f|_{[x+\tfrac{1}{n},\infty)} \ $. These sequences satisfy the conditions of the MCT and so

$\lim_{n\to\infty}F(x-\tfrac{1}{n})=\lim \int_{-\infty}^x f_n(x) = F(x) \ $

$\lim \int f^*_n = \int_x^\infty f \implies \lim_{n\to\infty} F(x+\tfrac{1}{n})=F(x) \ $.

Therefore, $\forall \epsilon>0 \ \exists N : n\geq N \implies F(x)-F(x-\tfrac{1}{n})<\epsilon>F(x+\tfrac{1}{n})-F(x) \ $. If we set $\delta=1/N \ $, then this implies $|F(x)-F(y)|<\epsilon \ $ for $|x-y|<\delta \ $, the definition of continuous.

4.8 We aim to show 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 $.

Observe that by the definition of limit inferior, $ \lim \inf f_n = \sup_n \inf_{k\ge n} f_k $.

So define $ g_n = \inf_{k\ge n} f_k $. Note that $ g_n $ is a nonnegative measurable function, and observe that $ g_n\le f_n \implies \int g_n\le \int f_n $, and $ \lim g_n = \lim \inf f_n $.

We have

$ \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 $.

4.14

a.) We aim to show that, under the hypotheses of Theorem 17, we have $ \int \left|{f_n-f}\right| \to 0 $.

So let $g_n, f_n, g, f \ $ be as in the theorem and note $|f_n|\leq g_n \implies |f|\leq g \ $. Therefore, $|f-f_n|\leq |f_n|+|f|\leq g_n+g \ $, and so $ h_n=g_n+g-|f_n-f| \geq 0 \ $ is a sequence of nonnegative measurable functions.

By Fatou, we have $\int \lim h_n \leq \lim \inf \int h_n \ $,

which becomes

$\int 2g \leq \int 2g - \lim \sup \int |f_n-f| \ $.

This gives $\lim \sup \int |f_n-f| \leq 0 \leq \lim \inf \int |f_n-f| \implies |f_n-f|\to 0 \ $.

b.) Suppose $ \left\{{f_n}\right\} \ $ is a sequence of integrable functions such that $ f_n \to f \ $ a.e. with $ f \ $ integrable. We aim to show that $ \int \left|{f-f_n}\right| \to 0 \iff \int \left|{f_n}\right| \to \int \left|{f}\right| \ $.

Since the $\Leftarrow \ $ follows from part (a) trivially, examine the other direction and observe that

$\int \left|{f-f_n}\right| \to 0\implies |\int | f_n|-\int|f||\leq \int ||f_n|-|f||\leq \int|f_n-f|\to 0 \implies \int|f_n|\to\int |f| \ $.

4.15

a.) Let $ f \ $ be integrable over $ E \ $. Then, given $ \epsilon > 0 $, we aim to show there is a simple function $ \phi \ $ such that $ \int_E \left|{f-\phi }\right| < \epsilon \ $.

By problem 4, there exists a simple function $ \phi_1 \le f \ $ such that $ \int_E f - \frac{\epsilon}{2} < \int_E \phi_1  \ $ for $ f^+ \ $. Similarly, there is a simple function $ \phi_2 \le f $ such that $ \int_E f - \frac{\epsilon}{2} < \int_E \phi_2 $ for $ 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{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon $.

Therefore, $ \int_E \left|{f-\phi }\right| < \epsilon $.

b.) Under the same hypothesis, we aim to show there is a step function $ \psi $ such that'' $ \int_E \left|{f-\psi }\right| < \epsilon $.

Set $f_n = f|_{[-n,n} \ $. Observe $|f_n|\leq |f|, \ f_n\to f \ $. By the dominated convergence theorem,

$\int |f-f_n|\to 0 \ $.

which implies

$\int |f|_{(-\infty,-n)\cup(n,\infty)} |\to 0 \ $,

so there is an $N \ $ such that $\int |f|_{(-\infty,-N)\cup(N,\infty)} |<\epsilon \ $. By part a, then, there is a simple function $\phi \ $ such that

$\int |f-\phi |<\epsilon \ $.

By Proposition 3.22, we have a step function $\psi \ $ such that...

c.) Under the same hypothesis, we aim to show there is a continuous function $ g $ vanishing outside a finite interval such that'' $ \int_E \left|{f-g}\right| < \epsilon $.

Note that by the previous part, we have a step function $\psi \ $ such that

$\int |f-\psi|<\epsilon/2 \ $.

As in a previous homework problem, we may "continuousize" this function outside a set of measure $\epsilon/2 \ $. Call this continuousization $g \ $.

Then $\int |f-g| \leq \int |f-\psi| + \int |\psi-g| < \epsilon \ $.

16. We aim to establish the Riemann-Lebesgue Theorem, which states that if $ f \ $ is an integrable function on $ (-\infty, \infty ) \ $, then $ \lim_{n\to \infty} \int_{-\infty}^\infty f(x) \cos nx dx = 0 \ $.

Let $ f \ $ be an integrable function on $ (-\infty, \infty ) $ and let $ \epsilon > 0 $.

Define a step function $ \psi \ $ such that $ \int_E \left|{f-\psi }\right| < \frac{\epsilon}{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{\epsilon }{2} + \int \left|{\psi (x) \cos nx }\right|dx $.

As $ n\to \infty $ we have

$ \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{\epsilon}{2} $. Therefore, $ \left|{ \int f(x) \cos nx dx}\right| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon $

and

$ \lim_{n\to \infty} \int f(x) \cos nx dx = 0 $.

Final Question:

We are asked to find an example of a sequence $ \left\{{f_n}\right\} $ of bounded functions on [0,1] that tends to 0 in measure, but does not converge anywhere, i.e. find a sequence of functions on [0,1] that converges in measure but does not converge pointwise anywhere.

So define the sets $S_{mn} = [\tfrac{m}{n},\tfrac{m+1}{n}] \ $, where $0\leq m 0 }\right\} = \lim mT_k = 0 \ $, but we also note that for any point $x \ $ and any $N\in\mathbb{N} \ $, there will be a $S_{mN} \ $ such that $x\in S_{mN} \ $. Hence, $f_n \ $ cannot converge pointwise anywhere, because for any integer, there will be $T_k \ $ past that integer containing x, and $T_k \ $ not containing x. So at any point x, $f_n(x) \ $ will continue to oscillate between 0 and 1 infinitely often as $n\to\infty \ $.