### Lebesgue Measure and Outer Measure Problems

More proving, still on Real Analysis. This is my solution and if you find any errors, do let me know.

### Problems

Lebesgue Measure: Let $\mu$ be set function defined for all set in $\sigma$-algebra $\mathscr{F}$ with values in $[0,\infty]$. Assume $\mu$ is countably additive over countable disjoint collections of sets in $\mathscr{F}$.
1. Prove that if $A$ and $B$ are two sets in $\mathscr{F}$, with $A\subseteq B$, then $\mu(A)\leq \mu(B)$. This property is called monotonicity.
2. Prove that if there is a set $A$ in the collection $\mathscr{F}$ for which $\mu(A)<\infty$, then $\mu(\emptyset)=0$.
3. Let $\{E_{k}\}_{k=1}^{\infty}$ be a countable collection of sets in $\mathscr{F}$. Prove that $\mu\left(\displaystyle\bigcup_{k=1}^{\infty}E_{k}\right)\leq \displaystyle\sum_{k=1}^{\infty}\mu(E_k)$
Lebesgue Outer Measure:
1. By using property of outer measure, prove that the interval $[0,1]$ is not countable.
2. Let $A$ be the set of irrational numbers in the interval $[0,1]$. Prove that $\mu^{*}(A)=1$.
3. Let $B$ be the set of rational numbers in the interval $[0,1]$, and let $\{I_k\}_{k=1}^{n}$ be finite collection of open intervals that covers $B$. Prove that $\displaystyle\sum_{k=1}^{n}\mu^{*}(I_k)\geq 1$.
4. Prove that if $\mu^{*}(A)=0$, then $\mu^{*}(A\cup B)=\mu^{*}(B).$

### Solutions

1. Proof. If $A\subseteq B$, then $B= A\cup (B\cap A^c)\Rightarrow B= A\cup (B\backslash A)$. Thus, \nonumber \begin{aligned} \mu(B)&= \mu(A\cup (B\backslash A))\\ &= \mu(A)+\mu(B\backslash A)\\ &(\text{since}\;\mu\;\text{is countably additive on disjoint sets}) \end{aligned} We can see that $\mu(B)\geq \mu(A)$ since $\mu(B\backslash A) > 0$. $\hspace{13.5cm}\blacksquare$
2. Proof. For any set $A$ in $\mathscr{F}$ such that $\mu(A)<\infty$, $A\cup \emptyset = A$. Thus, \nonumber \begin{aligned} \mu(A)&=\mu(A\cup \emptyset)=\mu(A)-\mu(\emptyset)\\ 0&=\mu(\emptyset) \end{aligned} $\hspace{13.5cm}\blacksquare$
3. Proof. We define a sequence $\{A_n\}_{n=1}^{\infty}\subseteq\mathscr{F}$, such that $A_1=E_1$ and $$\nonumber A_n = E_n \backslash \bigcup_{k=1}^{n-1}E_k,\;\text{for}\;n>1$$ It is easy to see that $A_n$ is pairwise disjoint, and $\bigcup_{n=1}^{\infty}A_n=\bigcup_{k=1}^{\infty}E_k$, also $\{A_n\}\subseteq \{E_k\}$. Thus by countably additive and monotonicity property of $\mu$, we have \nonumber \begin{aligned} \mu\left(\bigcup_{k=1}^{\infty}E_k\right)&=\mu\left(\bigcup_{n=1}^{\infty}A_n\right)\\ &=\sum_{n=1}^{\infty}\mu(A_n)\\ &\leq \sum_{k=1}^{\infty}\mu(E_k)\;(\text{by monotonicty}). \end{aligned} $\hspace{13.5cm}\blacksquare$
4. Proof. Let's prove this by contradiction, assume the interval $[0,1]$ is countable. Then we need to show that $\mu^{*}([0,1])=0$ for it to be countable. Now consider $\varepsilon >0$, such that $I = \{[\varepsilon - 0, 1 + \varepsilon]\}$ covers $[0,1]$. Then by property of outer measure that says, $\mu^{*}([a,b])$ is the length of $[a,b]$, we have $$\nonumber \mu^{*}([0,1]) = \inf\,\{\ell (I)\} = (1+\varepsilon) - (0-\varepsilon) = 1+2\varepsilon$$ This holds for each $\varepsilon >0$, thus $\mu^{*}([0,1])=1$ which is a contradiction.$\hspace{2.13cm}\blacksquare$
5. Proof. If $A=\{\mathbb{Q}^c\cap [0,1]\}$ is the set of irrational numbers in the interval $[0,1]$, then $A^c=\{\mathbb{Q}\cap [0,1]\}$ is the set of rational numbers in the interval $[0,1]$. Now consider the following, \nonumber \begin{aligned} \mu^{*}([0,1])&=\mu^{*}(A)+\mu^{*}(A^{c})\\ \mu^{*}(A)&=\mu^{*}([0,1]) - \mu^{*}(A^{c})\\ &=1 -\mu^{*}(A^{c}) \end{aligned} We need to show that $\mu^{*}(A^{c})$ has outer measure zero. To do that, let $a_1,a_2,\cdots\in A^{c}$. And for $\varepsilon > 0$, $\exists$ $\{I_n\}$ such that $\{I_n\} = \{(a_n - \frac{\varepsilon}{2^{k+1}}, a_n + \frac{\varepsilon}{2^{k+1}})\}$. Thus, $\bigcup_{n=1}^{\infty}I_n$ covers $A^{c}$, and by outer measure we have, \nonumber \begin{aligned} \mu^{*}(A^c)& \leq \inf\left\{\sum_{n=1}^{\infty}\ell(I_n)\right\}\\ &\leq \inf\left\{\sum_{n=1}^{\infty}\left(a_n + \frac{\varepsilon}{2^{k+1}} - a_n + \frac{\varepsilon}{2^{k+1}}\right) \right\}\\ &\leq \inf\left\{\sum_{n=1}^{\infty}\left(\frac{\varepsilon}{2^{k}}\right) \right\}\\ &\leq \varepsilon \end{aligned} Since this hold for each $\varepsilon$ then $\mu^{*}(A^c)=0$. Thus, $\mu^{*}(A)=1-0=1$.
$\hspace{13.5cm}\blacksquare$
6. Proof. The rational numbers are dense in $\mathbb{R}$. Thus, any point in the interval $[0,1]$ may it be irrational numbers will always be a point of closure of $B$, that is $\bar{B}=[0,1]$. Since $B\subseteq \bigcup_{k=1}^{\infty}I_k$, then by closure property, $\bar{B}\subseteq \overline{\bigcup_{k=1}^{\infty}I_k}$, which is $\bar{B}\subseteq \bigcup_{k=1}^{\infty}\bar{I}_k$. Thus by definition of outer measure we have, \nonumber \begin{aligned} 1=\mu^{*}([0,1])&=\mu^{*}(\bar{B})\leq \mu^{*}\left(\bigcup_{k=1}^{\infty}\bar{I}_k\right)\\ &\leq \sum_{k=1}^{\infty}\mu^{*}(\bar{I}_k)=\sum_{k=1}^{\infty}\mu^{*}(I_k). \end{aligned} Thus, $$\nonumber\sum_{k=1}^{\infty}\mu^{*}(I_k)\geq 1$$ $\hspace{13.5cm}\blacksquare$
7. Proof. We need to show that, \nonumber \begin{aligned} &\mu^{*}(A\cup B)\leq \mu^{*}(B)\\ &\mu^{*}(B)\leq \mu^{*}(A\cup B) \end{aligned}
1. From the definition of outer measure, \nonumber \begin{aligned} \mu^{*}(A\cup B)&\leq \mu^{*}(A)+\mu^{*}(B)\\ &\leq \mu^{*}(B). \end{aligned}
2. Since $B\subseteq A\cup B$, then from property of outer measure that if $A\subseteq B$, then $\mu^{*}(A)\leq \mu^{*}(B)$. Hence, $$\nonumber \mu^{*}(B)\leq \mu^{*}(A\cup B)$$
$\hspace{13.5cm}\blacksquare$