a characterization of $L^p$ space The following question should be part of the questions I recently asked here Prove or disprove a claim related to $L^p$ space
If $g \in L^p(\Omega, \lambda)$ where $\Omega$ is a bounded subset of $\mathbb{R^n}$, $p>1$ and $\lambda$ is the Lebesgue measure. By Holder's inequality, we know that for any measurable set $E \subset \Omega$, $$\int_E |g| d \lambda \le ||g||_p \lambda (E)^{\frac{p-1}{p}}$$.
Now the question is, if there exits a constant C, such that for any measurable set $E \subset \Omega$, $$\int_E |g| d \lambda \le C \lambda (E)^{\frac{p-1}{p}}$$ Does this imply $g \in L^p(\Omega, \lambda)$?

Here is my partial work. I tried to use the the duality of $L^q$ space by contradiction or the fact that simple functions are dense to prove this characterization, but I cannot control the constant. I think I need a powerful elementary inequality. Or maybe this characterization is not true. 
Any comments would be appreciated. Thanks!
 A: The condition does not imply that $g\in L^p$.
Consider $g(x)=x^{-1/p}$ on $[0,1]$. Obviously, $g(x)\not\in L^p([0,1])$. 
Let $E_\alpha=\{x:x\le\alpha^{-p}\}$, then $|E_\alpha|=\alpha^{-p}$ and
$$
\begin{align}
\int_{E_\alpha}g(x)\,\mathrm{d}x
&=\frac{p}{p-1}\alpha^{1-p}\\
&=\frac{p}{p-1}|E_\alpha|^{\frac{p-1}p}
\end{align}
$$
Since $g(x)$ is smaller on any other set of measure $\alpha^{-p}$, we have
$$
\int_Eg(x)\,\mathrm{d}x\le\frac{p}{p-1}|E|^{\frac{p-1}p}
$$
A: Indeed, let $h=\Sigma_ia_i\chi_{E_i}$ be a simple function with $E_i \subset \Omega, E_i \cap E_j = \emptyset$ and $||h||_{\frac{p}{p-1}}=1$. $$\int gh= \Sigma_ia_i\int _{E_i} g \le C \Sigma_i a_i |E_i|^{\frac{p-1}{p}}=C||h||_{\frac{p}{p-1}}=C $$
Hence by duality and the fact that simple functions are dense, we have $g \in L^p$ and $||g||_p \le C$.

Remark: I think you made a mistake when calculating  $||h||_{\frac{p}{p-1}}$. I made a mistake at the very beginning... You partial work is actually in the right direction!
A: Later my friends proved that the condition in the question holds if and only if $g$ is weak $L^p$. 
Here is the proof:
If $g$ satisfies the condition in the question, then let $E=\{|g|>\mu\}$, then we have $$\mu |E| \le \int_E |g| \le C|E|^{1-1/p}$$Hence $\mu |E|^{1/p} \le C$, thus $g$ is weak $L^p$.
If $g$ is weak $L^p$, then $\int_E |g|=\int_0^{\infty} |E \cap \{|g|>\mu\} d \mu \le t|E| + \int_t^{\infty}\frac{C}{\mu ^p} d \mu \le t|E|+Ct^{1-p}, \forall t>0$. Hence the condition follows by letting $t=|E|^{-1/p}$.
In particular, the proof shows there exists $g \in L^{p,w}-L^p$ such that $g$ satisfies the condition in the question. 
