User:J D Bowen/Math710 HW4

3.20) We aim to show $$\chi_{A\cap B}=\chi_A \cdot \chi_B \ $$.


 * 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: $$x\in A \ \and \ x\not\in B \ $$.

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

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

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


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

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 4: $$x\in A \ \and \ x\in B \ $$.

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

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

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

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

23) We aim to show the following 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 $$ \epsilon >0\ $$, we can find a step function $$ g\ $$ and a continuous function $$ h\ $$ such that

$$ \left|{f-g}\right| < \epsilon $$ and $$ \left|{f-h}\right| < \epsilon $$,

except on a set of measure less than $$ \epsilon $$; i.e.,

$$ m \left\{{ x: \left|{f(x)-g(x)}\right| \ge \epsilon }\right\} < \epsilon $$

and

$$ m \left\{{ x: \left|{f(x)-h(x)}\right| \ge \epsilon }\right\} < \epsilon $$.

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 $$ \epsilon > 0 $$, we aim to show there is an $$ M \ $$ such that $$ \left|{f}\right| \le M $$ except on a set of measure less than $$ \frac{\epsilon}{3} $$.

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. Hence the sequence $$mE_n \ $$ has a limit. Since $$f \ $$ is only $$\pm \infty \ $$ on a set of measure zero, that limit is zero. Now we 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 $$ \epsilon > 0 \ $$ and $$ M\ $$, we aim to show there is a simple function $$ \varphi $$ such that $$ \left|{f(x)-\varphi (x)}\right| < \epsilon $$ 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 such that $$M/N<\epsilon \ $$. For $$k\in\left\{{-N, \dots, N-1}\right\} \ $$, let $$E_k=\left\{{x\in[a,b]: kM/N \leq f(x)<(k+1)M/N }\right\} \ $$. Note that since $$f \ $$ is measurable, so are all the $$E_k \ $$. Define

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

Note that 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 argument works just as well if we forbid $$|k|<m \ $$.

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

Let $$\varphi=\sum_{i=1}^n a_i \chi_{A_i} \ $$.

Let $$I_{i,j} \ $$ be an open cover of $$A_i \ $$ such that $$m(U_i \Delta A_i)<\epsilon/3n \ $$.

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 use the same trick as before to restrict the result to $$m\leq g \leq M \ $$

d) Given a step function $$ g $$ on $$ [a,b] $$, we aim to show there is a continuous function $$ h $$ such that $$ g(x) = h(x) $$ except on a set of measure less than $$ \frac{\epsilon}{3} $$. 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 $$(x_i-\tfrac{\epsilon}{2n},x_i+\tfrac{\epsilon}{2n}) \ $$.

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 (hence continuous) and satisfies $$h_i(x_i-\tfrac{\epsilon}{2n})=g(x_i-\tfrac{\epsilon}{2n}), h_i(x_i+\tfrac{\epsilon}{2n})=g(x_i+\tfrac{\epsilon}{2n}) \ $$.

So if we 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) We aim to show that if $$ f \ $$ is measurable and $$ B \ $$ is a Borel set, then $$ f^{-1} (B) \ $$ is a measurable set - with the hint that 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.


 * 1) Closed under complementation:
 * Suppose $$E\in C \ $$. Then we have $$f^{-1}(\overline{E})=\overline{f^{-1}(E)} \ $$, which is the complement of a measurable set and hence measurable, and so $$\overline{E}\in C \ $$.


 * 2) 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 hence measurable. So $$\cup_{j=1}^\infty E_j \in C $$.

Hence, $$C \ $$ is a $$\sigma \ $$-algebra.

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

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.

Note that $$g \ $$ is automatically measurable. Then using the previous problem, observe that $$\left\{{x:g\circ f (x)>r\in\mathbb{R} }\right\} = (g\circ f)^{-1}(r,\infty) = f^{-1}(g^{-1}(r,\infty)) \ $$ which is measurable.

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

a) We aim to show that $$ f $$ is a homeomorphism of $$ [0,1] $$ onto $$ [0,2] $$.

Observe that $$x\mapsto f_1(x) \ $$ and $$x\mapsto x \ $$ are both continuous functions, and so their sum $$f \ $$ is continuous. We remember from the 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 a bijection. Finally, since $$f \ $$ is strictly increasing and continuous, it admits a continuous inverse. Hence, $$f \ $$ is a homeomorphism.

b) We aim to show that $$ f $$ maps the Cantor set onto a set $$ F $$ of measure 1. 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 $$ g = f^{-1} \ $$. We aim to show that there is a measurable set $$ A \ $$ such that $$ g^{-1}(A) \ $$ is not measurable.  By part (b), we know there exists a set $$B \subset [0,2] \ $$ such that $$g(B)=C \ $$, the Cantor set and $$mB=1 \ $$.  Since there exists an unmeasurable set within any set of positive measure, there is a subset $$V\subset B \ $$ that is unmeasurable.  Then $$m(g(V)) \leq m(g(B))=m(C)=0 \ $$, and so $$g(V) \ $$ is measurable.  Define $$A=g(V) \ $$; this set satisfies the conditions we seek.

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

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