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\bigcap_{i \mathop= n}^\infty E_i $ and:
 * $G_n=F_n\setminus F_{n-1},$

with $F_1=G_1$.

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

By the definition of a $\sigma$-algebra, $F_n$ and $G_n$ are measurable for every $n$.

Because measure is monotone, $\mu(G_n)\leq\mu(F_n)\leq\mu(E_n)$ for every $n$.

The Sequence $\left\{\mu(F_n)\right\}_{n=1}^\infty$ has a limit
If $x\in F_n$ for some $n$, then $x\in\displaystyle\cap_{i \mathop= n}^\infty E_i\subseteq\cap_{i \mathop = n+1}^\infty E_i=F_{n+1}$,

so $F_n\subseteq F_{n+1}$ for every $n$.

Using the additivity of the measure again, we have $\mu(F_n)\leq\mu(F_{n+1})$ for every $n$, and the sequence $\left\{F_n\right\}_{n=1}^\infty$ is nondecreasing.

By the Monotone Convergence Theorem for sequences, the sequence does have limit. (The limit may be $\infty$).

The Sum $\displaystyle\sum_{n \mathop= 1}^\infty \mu(G_n)=\lim_{n\rightarrow\infty}\mu(F_n)$
$G_1=F_1\setminus F_0=F_1\setminus\varnothing=F_1$.

If $F_n=\displaystyle\cup_{i\mathop= 1}^n G_i$ for some $i$, then

Therefore, by induction, $F_n=\cup_{i=1}^nG_i$ for all $n$.

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

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

If $n>n_1$, then $x\not\in G_n$ since $x\in F_{n_1}$.

The sets $G_n$ are therefore disjoint since a point $x$ can be an element of at most one of them.

Because of this and because $F_n=\cup_{i \mathop= 1}^nG_n$ for every $n$, we have:
 * $\mu(F_n)=\displaystyle\sum_{i \mathop 1}^n\mu(G_i)$

for every $n$ and:
 * $\displaystyle\lim_{n\rightarrow\infty}\mu(F_n)=\sum_{n \mathop 1}^\infty\mu(G_n)$

Since $F_n\subseteq F_{n+1}$ for every $n$, we have $F_n=\displaystyle\cup_{i\mathop= 1}^nF_i$ for every $n$.

Using the above work,

Since $\mu(F_n)\leq\mu(E_n)$ for every $n$, we have:
 * $\mu\left({\lim\inf E_n}\right)=\displaystyle\lim_{n\rightarrow\infty}\mu\left({F_n}\right)\leq\lim\inf\mu\left({E_n}\right)$

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:
 * $x\in\displaystyle\bigcap_{i \mathop= n}^\infty E_i \subseteq \bigcup_{n \mathop= 1}^\infty \bigcap_{i \mathop= n}^\infty E_i = \lim\inf E_n$.