Help with Proving This Lemma 
Lemma Let $\Omega$ be a $\sigma$-finite measure space and $J \colon \Omega \to [0, \infty]$ a measurable function. If $1 < p < \infty$ and $F \ge 0$ then the following statements are equivalent:
  
  
*
  
*$\lVert J \rVert_p \le F$; 
  
*$\forall g \in L^{p'}(\Omega), g \ge 0, \int_{\Omega} g^{p'}dx \le 1$ we have $\int_{\Omega}Jg\, dx \le F$, where $p^\prime$ is the conjugate of $p$.
  

This is a lemma from this ancient question(Infinite dimensional integral inequality), which originally came from a text of Hardy-Littlewood-Polya. The symbols in that text were too old to follow, and so I tried to prove it on my own. But I had some troubles.
The direction "$1\Rightarrow 2$" follows by Holder's inequality.
For the direction "$2\Rightarrow 1$", at first I wanted to use the Rieze representation theory, but later I found two flaws in my reasoning. And I can only prove a weaker version of this direction.
Let statement $2^\prime$ be that, $\forall g \in L^{p'}(\Omega),  \int_{\Omega} |g|^{p'}dx \le 1$ we have $\int_{\Omega}Jg\, dx \le F$, and further, $J\in L^p.$
My proof for "$2^\prime\Rightarrow 1$":
By statement $2^\prime$, we know $\int_{\Omega}Jg\, dx$ is a continuous functional on $L^{p^\prime}(\Omega)$ and with a operator norm less than or equal to $F$. Because its norm is just $||J||_p$, we have the statement $1$.
My question is how to prove the direction "$2\Rightarrow 1$".
Thanks for help.
 A: Let $(\Omega,\mathcal A, \mu)$ be the measure space in question. Since $\mu$ is a $\sigma$-finite measure there exist $C_n$ measurable sets s.t. $\mu(C_n)<\infty$, $C_1\subset C_2 \subset \ldots$ and $\bigcup_n C_n = \Omega$.
First we will show that $\mu(\{J=\infty\})=0$. For arbitrary $K>0$ and $n$ define $r_{n,K}(\omega) :=  I(\{J> K\} \cap C_n) $. Then for any $q>1$ we have $\int r_{n,K}^q d\mu< \mu(C_n)<\infty$, hence one can define $s_{n,K}(\omega):= r_{n,K}(\omega)^{p-1} / (\int r_{n,K}^{p})^{(p-1)/p} = I(C_n\cap \{J>K\}) / \mu(C_n \cap \{J>K\})^{(p-1)/p}$. Since $\int s_{n,K}^{p'} d\mu =1$ and $r_{n,k} \ge 0$, we can use the assumptions
$$ F\ge \int s_{n,K} J d\mu \ge K \mu(C_n\cap \{J>K\})^{1/p}, $$
hence $\mu(C_n\cap \{J>K\})\to 0$ as $K\to \infty$ for any $n$. Thus taking $n\to \infty$ we obtain $\mu(\{J=\infty\})=0$, using the continuity of the measure. 
Next we will show that $J\in L^p$ and that $||J||_p\le F$ holds. Define $h_{n,K}:= J\cdot I(\{J\le K\} \cap C_n)$. Due to the fact that $\int h_{n,K} ^{q} \le K^{q} \mu(C_n)<\infty $ for any $q>1$, we can define $g_{n,K} := (h_{n,K}^{p-1}) / (\int h_{n,K}^{p}d\mu))^{(p-1)/p}$. Now $g_{n,K} \ge 0$ and $\int g_{n,K}^{p'} d\mu = 1$, hence the assumptions are satisfied, so $$ \int J g_{n,K} d\mu \le F $$
for any $n$ and $K$. Denote $h_{n,n}$ by $h_{n}$. Then $h_{n}\ge0$ and $h_n\le h_{n+1}$ hence by the Beppo-Levi theorem
$$ \lim_n \int h_{n}^{q} d\mu \to  \int \lim_n h_n^{q}d\mu $$
for any $q>1$. 
 Let $g_n$ denote $g_{n,n}$. Using Fatou's lemma
$$ \int \liminf_n g_{n,n} Jd\mu \le\liminf_n \int g_{n,n}J d\mu = ||J||_p \le F  $$ 
