Bartle's Theorem 6.14 Completeness of $L^p$ (Riesz Representation)- need explanation Note: There are 4 parts to my questions, I have labelled (and colored them, sorry if that is irritating) with (1), (2), (3) and (4) of the proof. Any answer to parts is great. Thanks!  

Completeness Theorem. If $1 \le p < \infty$, then the space $L_p$ is a complete normed linear space under the norm 
  $$ ||f ||_p = \Big \{ \int |f|^p \, d \mu \Big \} ^{1/p}. $$ 
Proof: Let $(f_n)$ be a Cauchy sequence relative to the norm $|| \, \, || _p$. There exists a sub-sequence $(g_k)$ of $(f_n)$ such that $||g_{k+1} - g_k || _p < 2^{-k}$ for $k \in \mathbb{N}$. Define $g$ by 
  $$ g(x) = |g_1(x) | + \sum_{k=1}^{\infty} |g_{k+1}(x) - g_k(x) |$$
  so $g : X \rightarrow \bar{ \mathbb{R}}^{+}$ is measurable. By Fatou's Lemma, we have 
  $$ \color{blue}{ \int |g|^p \, d \mu \le \lim \inf _{n \rightarrow \infty} \int \Big \{ |g_1| + \sum_{k=1}^n |g_{k+1}- g_k| \Big \}^p \, d \mu \quad \quad (1) } $$
  Taking the $p^{th}$ root of both sides and apply Minkowski's Inequality to obtain, 
  $$ \color{blue} { \Big \{ \int |g|^p \, d \mu \Big \} ^{1/p } \le \lim \inf_{n \rightarrow \infty} \Big \{ ||g_1||_p + \sum_{k=1}^n ||g_{k+1} - g_k||_p \Big \} \quad (2) } $$
  $$ \le ||g_1||_p + 1. $$ 
  Hence, if $E = \{x \in X:g(x) < + \infty \},$ then $\color{blue}{ \text{  $E$ is a measurable set and $\mu(X \setminus E) = 0$. } \quad (3) } $ Therefore, $\color{blue}{ \text{ the series in $g(x)$ converges almost everywhere and $g\, \chi_E$ belongs to $L_p$. } \quad (4) } $


My questions regarding 
(1) : By construction $$g(x) = |g_1(x)| + \lim \sum_{k=1}^n |g_{k+1}(x) -  g_k(x)| $$ $$ =  |g_1(x) | + \lim \inf \sum_{k=1}^n |g_{k+1}(x) - g_k(x) |. $$ Is this right? 
(2) : If we take the $p^{th}$ root we would obtain, an exponent of $1/p$ outside of the $\lim \inf $. That is, from (1) we should obtain
$$ \Big( \lim \inf _{n \rightarrow \infty} \int \Big \{ |g_1| + \sum_{k=1}^n |g_{k+1}- g_k| \Big \}^p \, d \mu \Big) ^{1/p} $$ 
so we could not directly apply Minkowski's. Does this mean it's true, for a sequence $a_n$ and continuous function $f$ that $f ( \lim \inf a_n) \le \lim \inf f(a_n)$? 
(3) & (4) : It is unclear why these statements are true. 
Thanks in advance. 


Minkowski's Inequality. If $f$ and $h$ belong to $L_p$, $p \ge 1 $ then $f+h$ belongs to $L_p$ and 
  $$ ||f+h||_p \le ||f||_p + ||h||_p $$

 A: Addressing your questions as numbered in the OP:
(1) Yes, any series $\sum_{k=1}^{\infty} a_k$ of nonnegative terms is convergent in the extended reals, hence if $s_n = \sum_{k=1}^{n}a_k$ is the $n$th partial sum, then $\lim s_n = \liminf s_n = \limsup s_n$.
(2) If $a_n$ is a sequence of extended real numbers, we have that $b_n = \inf_{k \geq n}a_k$ is an increasing sequence, so it converges to its supremum, which is $\sup_{n \geq 1}b_n = \lim_{n \geq 1} b_n = \lim_{n \geq 1}\inf_{k \geq n}a_k = \liminf a_n$. Therefore,
$$f(\liminf a_n) = f(\lim_{n\geq 1} b_n)$$
If $f$ is continuous then we may pull the limit outside:
$$f(\lim_{n\geq 1} b_n) = \lim_{n \geq 1}f(b_n) = \lim_{n \geq 1}f(\inf_{k \geq n}a_k)$$
If $f$ is also monotonically increasing then since $\inf_{k \geq n}a_k \leq a_k$ for all $k \geq n$, it follows that $f(\inf_{k \geq n}a_k) \leq f(a_k)$ for all $k \geq n$, hence $f(\inf_{k \geq n}a_k) \leq \inf_{k \geq n}f(a_k)$. Combining this with the above allows us to conclude that $f(\liminf a_n) \leq \liminf f(a_n)$ if $f$ is continuous and monotonically increasing. Apply this to $f(x) = x^{1/p}$ for $x \geq 0$ to obtain (2).
(3) $g$ is the limit of a sequence of measurable functions, hence measurable. If $X \setminus E$ has positive measure then $|g|$ is infinite on a set of positive measure, hence $\int |g|^p$ is infinite, contradicting (2).
(4) The series in (1) converges (in the real numbers) precisely when $g(x) < \infty$, which is precisely on the set $E$. Since the complement $X \setminus E$ has zero measure, this means that (1) converges almost everywhere. Finally, $$\int_X |g(x)\chi_E(x)|^p\ dx = \int_X |g(x)|^p \chi_E(x)\ dx = \int_E |g(x)|^p\ dx \leq \int_X |g(x)|^p\ dx < \infty$$
hence $g\chi_E \in L^p$. The rightmost inequality holds because of (2).
