User:Kcbetancourt

7.4.37 A commutative ring $$ R\ $$ is called a local ring if it has a maximal ideal. Prove that if $$ R\ $$ is a local ring with maximal ideal $$ M\ $$, then every element of $$ R-M\ $$ is a unit.

Since $$M \ $$ is a maximal ideal, it is prime. Let $$x,y\in R-M $$. If we had $$xy\in M \ $$, then since $$M \ $$ is prime, we have $$x\in M \ $$ or $$y \in M \ $$. Hence, $$R-M \ $$ is closed under multiplication.

Let $$ x\in R-M $$. Consider the ideal $$ (x)\ $$ generated by $$ x\ $$. Then, since $$ M\ $$ is the maximal ideal, $$ (x)\subseteq M $$. But $$ x\notin M $$ so we must have $$ M\subseteq (x) $$. But again, since $$ M $$ is the maximal ideal, this means that $$ (x) = R $$. Since $$ 1 \in R, \exists y \in R : xy = 1 $$. This implies that $$ x\ $$ is a unit. Therefore, every element of $$ R-M\ $$ is a unit.

<= Let R be a commutative ring with 1 and the set of all non-units be M. We want to show that M is a Unique Maximal in R. Assume that M is not maximal. Then $$ \exists \ $$ an ideal, I, in R such that M $$ \subset \ $$ I $$ \subset \ $$ R. And M $$ \ne \ $$ I $$ \ne \ $$ R. We know that 1 does not exist in M, because 1 is a unit. So if M is not equal to I, then $$ \exists \ $$ u $$ \in \ $$ I, where u is a unit. Then 1 $$ \in \ $$ I. But that implies I=R. But if I=R, then I is not a maximal idea. If I is not maximal, then that implies that M is maximal. Now assume that M is not unique. Then $$ \exists \ $$ M' such that M' is a maximal ideal in R. Then M' $$ \subset \ $$ R and M $$ \ne \ $$ R. So 1 $$ \notin \ $$ M'. Thus M' is the set of all non-units. That implies that M'=M. Therefore M is a unique maximal ideal in R.

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

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

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

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

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

$$ \chi _{A\cap B}(x) = \begin{cases} 1, & x\in A\cap B \\ 0, & x\notin A\cap B \end{cases} \ $$

Let $$ x\in A\cap B $$

$$ \chi _{A\cap B}(x) = 1 \iff x\in A $$ and $$ x\in B $$

$$ x\in A \iff \chi _A(x) = 1 $$

$$ x\in B \iff \chi _B(x) = 1 $$

So, $$ \chi _{A\cap B}(x) = 1 = 1 \cdot 1 = \chi _A(x) \cdot \chi _B(x) $$

Therefore, when $$ x\in A\cap B $$, $$ \chi _{A\cap B} = \chi _A \cdot \chi _B $$.

Now let $$ x\notin A\cap B $$

$$ \chi _{A\cap B}(x) = 0 \iff x\in A\setminus B $$ or $$ x\in B\setminus A $$

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 $$ \frac{\varepsilon}{3} $$.

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 $$ m\le \varphi \le M $$.

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 $$ m\ge g\ge M $$.

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 $$ m\ge h\ge M $$.

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.)

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.

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 homeomorphism of $$ [0,1] $$ onto $$ [0,2] $$.

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

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

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.