Characterization of $L^p$ function for $p>1$

Question

Let $\Omega$ be a bounded open set in $\mathbb{R}^n$, $1<p<\infty, 1/p +1/q=1,f \in L^1(\Omega)$.

Is it true that if there exists $C>0$ such that $sup \{\int fg: g$ is a step function, $ ||g||_q \le 1\}$$\le C$, then $f \in L^p(\Omega)$? (By step function, I mean the form $\Sigma_{i=1}^m a_i \chi_{Q_i}(x) $ where each $Q_i$ is a cube. )

I know that if replacing $g$ with simple functions above, then the conclusion is true. But I'm not sure whether it's also true for step functions. I guess it is true, because measurable functions can be approximated by continuous functions, and continuous functions can be approximated by step functions, in the sense of almost everywhere convergence or $L^s(\Omega)$ convergence, $\forall s \ge 1$.

However, by using this and some other corollaries, I find a paradox, so I'm confused about this result. Maybe it is not true. Can anyone confirm me whether this result is true or not?

Any comments would be appreciated.

The paradox is, I can claim if $f \notin L^p$, then for any $m>0$, there exists a cube $Q_m$ such that $$\int_{Q_m}f > m |Q_m|^{\frac{p-1}{p}} \quad \quad\quad\quad\quad\quad\quad\quad(*)$$ This can be proved by contradiction using the characterization of $L^p$ above. Now suppose $f(x)=1/|x|$ defined on $B_1 \subset \mathbb{R}^n, n \ge 3$, thus $f \notin L^n(B_1)$, then one can prove that there exists a universal constant $C(n)$ such that for any cube $Q$, we have $\int_Q 1/|x| \le C(n)|Q|^{\frac{n-1}{n}}$, contradicting $(*)$ if $p=n$

@TZakrevskiy, I've posted the paradox. I just don't understand why the duality theorem for simple functions are true, but for step functions are not. As you said, both of the set of simple functions and the set of step functions form a dense set. — student, Apr 02 '15 at 21:32
In the discussion of the Riesz representation theorem in Royden and Fitzpatrick, they explicitly construct a $g$ of $L^q$ norm $1$ that attains the bound given by Holder's inequality for a given $f$. Since simple functions are dense in $L^q$ you can approximate this $g$ by simple functions to get your result. — Ian, Apr 02 '15 at 21:35
Your statement is true for $1\lt p\lt\infty$, but $L^1$ and $L^\infty$ are not duals, as your counterexample shows. — robjohn, Apr 02 '15 at 21:51
@robjohn, in my example, $n$ can be chosen between 1 and $\infty$. I'm sorry I don't know what you mean. — student, Apr 02 '15 at 21:55
@Ian, yes, and I think the same argument holds for step functions. However, see the example I gave in my post... — student, Apr 02 '15 at 21:56
@Ian could you elaborate your comment? I don't see how the existence of such $g$ for a given $f\in L^1$ implies that $f\in L^p$. — TZakrevskiy, Apr 02 '15 at 21:56
@QinfengLi: The contrapositive of your statement is "If $f\not\in L^p$, then for any $C$, there is a step function $g$ so that $|g|_q\le1$ and $\int f(x)g(x),\mathrm{d}x\ge C$". Does this imply your statement about a single cube? — robjohn, Apr 02 '15 at 23:52
@robjohn, yes. See the argument here http://math.stackexchange.com/questions/1134176/a-characterization-of-lp-space — student, Apr 03 '15 at 01:00
@QinfengLi: the currently accepted answer there does not appear correct. — robjohn, Apr 03 '15 at 09:24
Moreover we clearly have $\int_Q\frac{1}{|x|},dx=\infty$ if $Q$ is centered at $0$ (because it contains a ball and there we can use polar coordinates..), so the statement about the universal constant is false! — Mizar, Apr 04 '15 at 12:17

Characterization of $L^p$ function for $p>1$

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