Monday, 8 September 2014

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, \begin{equation}\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} \end{equation} 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, \begin{equation}\nonumber \begin{aligned} \mu(A)&=\mu(A\cup \emptyset)=\mu(A)-\mu(\emptyset)\\ 0&=\mu(\emptyset) \end{aligned} \end{equation} $\hspace{13.5cm}\blacksquare$
  3. Proof. We define a sequence $\{A_n\}_{n=1}^{\infty}\subseteq\mathscr{F}$, such that $A_1=E_1$ and \begin{equation}\nonumber A_n = E_n \backslash \bigcup_{k=1}^{n-1}E_k,\;\text{for}\;n>1 \end{equation} 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 \begin{equation}\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} \end{equation} $\hspace{13.5cm}\blacksquare$
  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 \begin{equation}\nonumber \mu^{*}([0,1]) = \inf\,\{\ell (I)\} = (1+\varepsilon) - (0-\varepsilon) = 1+2\varepsilon \end{equation} 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, \begin{equation}\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} \end{equation} 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, \begin{equation}\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} \end{equation} 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, \begin{equation}\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} \end{equation} Thus, \begin{equation}\nonumber\sum_{k=1}^{\infty}\mu^{*}(I_k)\geq 1 \end{equation} $\hspace{13.5cm}\blacksquare$
  7. Proof. We need to show that, \begin{equation}\nonumber \begin{aligned} &\mu^{*}(A\cup B)\leq \mu^{*}(B)\\ &\mu^{*}(B)\leq \mu^{*}(A\cup B) \end{aligned} \end{equation}
    1. From the definition of outer measure, \begin{equation} \nonumber \begin{aligned} \mu^{*}(A\cup B)&\leq \mu^{*}(A)+\mu^{*}(B)\\ &\leq \mu^{*}(B). \end{aligned} \end{equation}
    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, \begin{equation}\nonumber \mu^{*}(B)\leq \mu^{*}(A\cup B) \end{equation}
    $\hspace{13.5cm}\blacksquare$

Reference

  1. Royden, H.L. and Fitzpatrick, P.M. (2010). Real Analysis. Pearson Education, Inc.

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