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 \ \land \ 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 \ \land \ 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 \ \land \ 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 \ \land \ 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) \lor 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.