User:Michellepoliseno

Analysis:

3.20. Show that the sum and product of two simple functions are simple. Show that:

$$ \chi _{A\cap B} = \chi _A \cdot \chi _B $$

$$ \chi _{A\cup B} = \chi _A + \chi _B - \chi _A \cdot \chi _B $$

$$ \chi _{A^c} = 1 - \chi _A $$.

Lets look at the cases: Case 1: $$x\not \in A \ \and \ x\not\in B \ $$.

This implies $$x\not \in A \cap B \ $$ and $$x\not\in A\cup B \ $$. We have

$$\chi_{A\cap B}=0 = 0\cdot 0 = \chi_A(x)\cdot \chi_B(x) \ $$,

$$\chi_{A\cup B}=0 = \ $$

Case 2: $$x\not\in A \ \and \ x\in B \ $$.

This implies $$x\not\in A \cap B \ $$ and $$x\in A\cup B \ $$. We have

$$\chi_{A\cap B}=0 = 0\cdot 1 = \chi_A(x)\cdot \chi_B(x) \ $$.

$$\chi_{A\cup B}=1 = 0+1-0= \chi_B(x)+\chi_A(x)-\chi_{A\cap B}(x)\ $$

Case 3: $$x\in A \ \and \ x\not\in B \ $$.

This implies $$x\not \in A \cap B \ $$ and $$x\in A\cup B \ $$. We have

$$\chi_{A\cap B}=0 = 1\cdot 0 = \chi_A(x)\cdot \chi_B(x) \ $$.

$$\chi_{A\cup B}=1 = 1+0-0 = \chi_B(x)+\chi_A(x)-\chi_{A\cap B}(x) \ $$

Now, Case 4: $$x\in A \ \and \ x\in B \ $$.

This implies $$x \in A \cap B \ $$ and $$x\in A\cup B \ $$. We have

$$\chi_{A\cap B}=1 = 1\cdot 1 = \chi_A(x)\cdot \chi_B(x) \ $$.

$$\chi_{A\cup B}=1 =1+1-1= \chi_B(x)+\chi_A(x)-\chi_{A\cap B}(x) \ $$

Therefore, note that $$x\in A \ \text{xor} \ x\not\in A \ $$, so $$\chi_A(x)+\chi_{\overline{A}}(x)=1 \ $$.

3.23. Prove Proposition 22 by establishing the following lemmas:

Proposition 22: Let $$ f\ $$ be a measurable function defined on an interval $$ [a,b]\ $$, and assume that $$ f\ $$ takes the values $$ \pm \infty $$ only on a set of measure zero. Then given $$ \varepsilon >0\ $$, we can find a step function $$ g\ $$ and a continuous function $$ h\ $$ such that $$ \left|{f-g}\right| < \varepsilon $$ and $$ \left|{f-h}\right| < \varepsilon $$ except on a set of measure less than $$ \varepsilon $$; i.e., $$ m \left\{{ x: \left|{f(x)-g(x)}\right| \ge \varepsilon }\right\} < \varepsilon $$ and $$ m \left\{{ x: \left|{f(x)-h(x)}\right| \ge \varepsilon }\right\} < \varepsilon $$. If in addition $$ m\le f\le M $$, then we may choose the functions $$ g\ $$ and $$ h\ $$ so that $$ m\le g\le M $$ and $$ m\le h\le M $$.

a.) Given a measurable function $$ f\ $$ on $$ [a,b]\ $$ that takes the values $$ \pm \infty $$ only on a set of measure zero, and given $$ \varepsilon > 0 $$, there is an $$ M\ $$ such that $$ \left|{f}\right| \le M $$ except on a set of measure less than $$ \frac{\varepsilon}{3} $$.

Consider the sets $$E_n = \left\{{x\in[a,b] : |f(x)|>n }\right\} \ $$. Clearly $$mE_0 \leq b-a \ $$, and since $$E_{n+1}\subseteq E_n \ $$, the sequence $$mE_n \ $$ is non-increasing. Thus the sequence $$mE_n \ $$ has a limit. Since $$f \ $$ is only $$\pm \infty \ $$ on a set of measure zero, that limit is zero. Now just apply the definition of a limit, and observe that for any $$\epsilon/3>0 \ $$, there is an $$M \ $$ such that $$mE_m < \epsilon/3 \ $$ whenever $$m>M \ $$.

b.) Let $$ f\ $$ be a measurable function on $$ [a,b]\ $$. Given $$ \varepsilon > 0 $$ and $$ M\ $$, there is a simple function $$ \varphi $$ such that $$ \left|{f(x)-\varphi (x)}\right| < \varepsilon $$ except where $$ \left|{f(x)}\right| \ge M $$. If $$ m\le f\le M $$, then we may take $$ \varphi $$ so that $$ m\le \varphi \le M $$.

Let $$N \ $$ be s.t. $$M/N<\epsilon \ $$. For $$k\in\left\{{-N, \dots, N-1}\right\} \ $$, and let $$E_k=\left\{{x\in[a,b]: kM/N \leq f(x)<(k+1)M/N }\right\} \ $$. Note since $$f \ $$ is measurable, so are all the $$E_k \ $$. Now, define

$$\varphi(x)=\sum_{k=-N}^{N-1} kM/N \chi_{E_k}(x) \ $$.

Note this simple function satisfies the requirements of the problem, since $$|f(x)|\leq M \implies \exists k: x\in E_k \implies |f(x)-\varphi(x)|<\epsilon \ $$, since all the bands are less than epsilon wide. This works as well if we forbid $$|k|<m \ $$.

c.) Given a simple function $$ \varphi $$ on $$ [a,b]\ $$, there is a step function $$ g\ $$ on $$ [a,b]\ $$ such that $$ g(x) = \varphi (x) $$ except on a set of measure less than $$ \frac{\varepsilon}{3} $$. If $$ m\ge \varphi \ge M $$, then we can take $$ g\ $$ so that $$ m\ge g\ge M $$.

Let $$\varphi=\sum_{i=1}^n a_i \chi_{A_i} \ $$, and let $$I_{i,j} \ $$ be an open cover of $$A_i \ $$ s.t. $$m(U_i \Delta A_i)<\epsilon/3n \ $$.

Now, define $$g=\sum_{i=1}^n a_i \chi_{U_i \backslash (U_1 \cup \dots \cup U_{i-1})} \ $$.

Note that $$g \ $$ is a step function and we have

$$g(x)\neq \varphi(x) \implies (g(x)=a_i\neq \varphi(x) \or g(x)=0\neq\varphi(x)=a_i \ $$

$$\implies x \in \bigcup_{i=1}^n U_i \Delta A_i, \ m \left({\bigcup_{i=1}^n U_i \Delta A_i}\right)<n\epsilon/(3n)=\epsilon/3 \ $$. We may restrict the result to $$m\leq g \leq M \ $$, using the same strategy as used earlier.

d.) Given a step function $$ g $$ on $$ [a,b] $$, there is a continuous function $$ h $$ such that $$ g(x) = h(x) $$ except on a set of measure less than $$ \frac{\varepsilon}{3} $$. If $$ m\ge g\ge M $$, then we may take $$ h $$ so that $$ m\ge h\ge M $$.

Let $$\left\{{x_0, \dots, x_n }\right\} \ $$ be a partition consisting of points where $$g \ $$ changes values. Given $$\epsilon \ $$, define the interval $$(x_i-\tfrac{\epsilon}{2n},x_i+\tfrac{\epsilon}{2n}) \ $$. Now, define $$m=\frac{g(x_i+\tfrac{\epsilon}{2n})-g(x_i-\tfrac{\epsilon}{2n})}{\epsilon/n} \ $$

and $$b=g(x_i-\tfrac{\epsilon}{2n})-m(x_i-\tfrac{\epsilon}{2n}) \ $$.

Then the function $$h_i:[x_i-\tfrac{\epsilon}{2n},x_i+\tfrac{\epsilon}{2n}]\to\mathbb{R} \ $$ is linear, thus continuous, and satisfies $$h_i(x_i-\tfrac{\epsilon}{2n})=g(x_i-\tfrac{\epsilon}{2n}), h_i(x_i+\tfrac{\epsilon}{2n})=g(x_i+\tfrac{\epsilon}{2n}) \ $$. Then define

$$h(x)=\begin{cases} g(x), & \mbox{if } \not\exists i:x\in[x_i-\tfrac{\epsilon}{2n},x_i+\tfrac{\epsilon}{2n}]  \\ h_i(x),  & \mbox{if } \exists i:x\in[x_i-\tfrac{\epsilon}{2n},x_i+\tfrac{\epsilon}{2n}] \end{cases} \ $$,

this function will be continuous and equal to $$g \ $$ except on a set of measure $$\epsilon \ $$. Changing $$\epsilon \to \epsilon/3 \ $$ changes nothing in this argument.

3.24. Let $$ f $$ be measurable and $$ B $$ a Borel set. Then $$ f^{-1} [B] $$ is a measurable set. (The class of sets for which $$ f^{-1} [E] $$ is measurable is a $$ \sigma $$-algebra.)

Let $$C \ $$ be the collection of sets for which $$ f^{-1} (E) \ $$ is measurable.

First show closed under complementation: Assume $$E\in C \ $$. Then $$f^{-1}(\overline{E})=\overline{f^{-1}(E)} \ $$, which is the complement of a measurable set and hence measurable, and so $$\overline{E}\in C \ $$.

Second show closed under countable unions: Let $$E_j, \ j=1, \dots \ $$ be a countable collection of sets in $$C \ $$. Then

$$f^{-1}\left({\bigcup_{j=1}^\infty E_j }\right) = \bigcup_{j=1}^\infty f^{-1}(E_j) \ $$, which is a countable union of measurable sets and thus is measurable. So $$\cup_{j=1}^\infty E_j \in C $$. Therefore, $$C \ $$ is a $$\sigma \ $$-algebra.

Now, observe that an open interval $$(a,b) \in C \ $$, since $$f \ $$ is measurable. Since $$C \ $$ is a $$\sigma \ $$-algebra containing the open intervals, it follows that the Borel sets are a subset of $$C \ $$.

3.25. Show that if $$ f $$ is a measurable real-valued function and $$ g $$ a continuous function defined on $$ (-\infty, \infty )\ $$, then $$ g \circ f $$ is measurable.

$$ (g \circ f) ^{-1} \ $$ $$ \implies \ $$ $$ f^{-1} (g^{-1} (-\infty, \infty )) \ $$, we know that $$ (g^{-1} (-\infty, \infty )\ $$ is measurable since g is continuous on $$ (-\infty , \infty )\ $$ . Then by 3.24, we know that since f is measurable, $$ f^{-1}\ $$ is measurable. Thus $$ (g \circ f) ^{-1} \ $$ is measurable.

3.28. Let $$ f_1 $$ be the Cantor ternary function, and define $$ f $$ by $$ f(x) = f_1(x) + x $$.

a.) Show that $$ f $$ is a homomorphism of $$ [0,1] $$ onto $$ [0,2] $$.

Note that $$x\mapsto f_1(x) \ $$ and $$x\mapsto x \ $$ are both continuous functions, and so their sum $$f \ $$ must be continuous. We remember from previous homework that $$f_1 \ $$ is non-decreasing, and since $$x\mapsto x \ $$ is strictly increasing, $$f \ $$ is strictly increasing. Since $$f(0)=0, f(1)=1+1=2 \ $$, and since it is continuous and strictly increasing, it is therefore a bijection. Thus, since $$f \ $$ is strictly increasing and continuous, it admits a continuous inverse. Therefore, $$f \ $$ is a homeomorphism.

b.) Show that $$ f $$ maps the Cantor set onto a set $$ F $$ of measure 1.

Note since $$f_1 \ $$ is constant on the complement of the Cantor set, and since that complement has measure one, the image of that set under the transform $$x\mapsto f_1(x)+x \ $$ is a set of measure 1. Since $$f([0,1])=[0,2] \ $$, which has measure two, the image of the complement of the complement of the Cantor set (ie, the Cantor set) must have measure 2-1=1.

c.) Let $$ g = f^{-1} $$. Show that there is a measurable set $$ A $$ such that $$ g^{-1}[A] $$ is not measurable.

Part b tells use that there exists a set $$B \subset [0,2] \ $$ s.t. $$g(B)=C \ $$, the Cantor set and $$mB=1 \ $$. Since there exists an unmeasurable set within any set of positive measure, there exists a subset $$V\subset B \ $$ which is unmeasurable. That implies that $$m(g(V)) \leq m(g(B))=m(C)=0 \ $$, and so $$g(V) \ $$ is measurable. Now, define $$A=g(V) \ $$; this set satisfies the conditions.

d.) Give an example of a continuous function $$ g $$ and a measurable function $$ h $$ such that $$ h \circ g $$ is not measurable. Compare with Problems 25 and 26.

Let $$g \ $$ be defined as earlier and $$h=\chi_A \ $$, then this is measurable, since $$A \ $$ is measurable.