User:Will.newton519/Sandbox

I am testing my ability to write for the proof of Fatou's Lemma for Measures. Feel free to comment. The proof currently is unfinished and needs to be revised. I will attempt to finish it later today.

Claim
Let $\{E_n\}_{n=1}^\infty$ be a sequence of measurable sets. Then:
 * $\mu\left({\lim\inf E_n}\right)\leq\lim\inf\mu\left({E_n}\right)$

Proof
For $n\in\N$, let:
 * $F_n=\displaystyle\left({

\bigcap_{i \mathop = n}^\infty E_i }\right) \setminus \left({ \bigcup_{i \mathop = 1}^{n-1} E_i }\right)$

By the definitions of Set Intersection and Set Difference, we have $F_n\subseteq E_n$ for every $n$.

By the definition of $\sigma$-algebra, $F_n$ is measurable for every $n$.

By Measure, $\mu(F_n)\leq\mu(E_n)$ for every $n$.

We now show that the sets $F_n$ are disjoint.

Suppose $x\in F_{n_1}$ for some $n_1$. Then $x\in E_{n_1}$.

If $n < n_1$, then $x\not\in F_n$ since $x$ cannot be in $E_n$.

If $n>n_1$, then $x\not\in F_n$ since $x\in E_{n_1}$.

The sets $F_n$ are therefore disjoint since a point $x$ can be an element of at most one of them. We next show that:
 * $\displaystyle\bigcup_{n \mathop= 1}^\infty F_n = \lim\inf E_n$

If $x\in\displaystyle\bigcup_{n \mathop = 1}^\infty F_n,$ then $x\in F_n$ for one $n$. This means:
 * $x\in\displaystyle\left({

\bigcap_{i \mathop = n}^\infty E_i }\right) \setminus \left({ \bigcup_{i \mathop = 1}^{n-1} E_i }\right)$ Therefore:
 * $x\in\displaystyle\bigcap_{i \mathop= n}^\infty E_i \subseteq \bigcup_{n \mathop= 1}^\infty \bigcap_{i \mathop= n}^\infty E_i = \lim\inf E_n$

If