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

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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 $\mathscr{F}$.

- 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*.
- Prove that if there is a set $A$ in the collection $\mathscr{F}$ for which $\mu(A)<\infty$, then $\mu(\emptyset)=0$.
- 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__:

- By using property of outer measure, prove that the interval $[0,1]$ is not countable.
- Let $A$ be the set of irrational numbers in the interval $[0,1]$. Prove that $\mu^{*}(A)=1$.
- 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 $\displaystyle\sum_{k=1}^{n}\mu^{*}(I_k)\geq 1$.
- Prove that if $\mu^{*}(A)=0$, then $\mu^{*}(A\cup B)=\mu^{*}(B).$