Is this a valid argument to show $|a|^q \in L^{(p/q)'}(\Omega)$? Brezis' functional analysis book includes a solution to one of its exercises that is looking somewhat fishy to me: while attempting to formalize the last step in the solution, I have arrived at a viciously circular argument. Have I misinterpreted Brezis' argument? Here is the exercise, its solution and my (mis?)interpretation of such solution (I have transcribed the first two myself, so any typos are probably mine) (You can ignore case 2 ($p = \infty$); the important step is the last step in case 1 ($p < \infty$)):
Exercise 4.7: Let $1 \leq q \leq p \leq \infty$. Let $a(x)$ be a measurable function on $\Omega$. Assume that $au \in L^q(\Omega)$ for every function $u \in L^p(\Omega)$. Prove that $a\in L^r(\Omega)$ with $r = \frac{pq}{p-q}$ if $p < \infty$ and $r = q$ if $p = \infty$ [Hint: Use the closed graph theorem.]
Solution: Consider the operator $T: L^p(\Omega)\to L^q(\Omega)$ defined by $Tu = au$. We claim that the graph of $T$ is closed. Indeed, let $(u_n)$ be a sequence in $L^p(\Omega)$ such that $u_n \to u$ in $L^p(\Omega)$ and $au_n \to f$ in $L^q(\Omega)$. Passing to a subsequence we may assume that $u_n \to u$ a.e. and $au_n \to f$ a.e. Thus $f = au$ a.e. and so $f = Tu$. It follows from the closed graph theorem (Theorem 2.9) that $T$ is bounded and so there is a constant $C$ such that
$(S1)$ $ \| au \|_q \leq C \| u \|_p$ $\forall u \in L^p(\Omega)$
Case 1: $p < \infty$. It follows from $(S1)$ that 
$\int |a|^q |v| \leq C^q \| v \|_{p/q}$ $\forall v \in L^{p/q}(\Omega)$
Therefore the map $v \mapsto \int |a|^q v$ is a continuous linear functional on $L^{p/q}(\Omega)$ and thus $|a|^q \in L^{ (p/q)' } $
Case 2: $p = \infty$. Choose $u = 1$ in $(S1)$.
How I interpreted the last step in the argument in case 1: from what I understand, when Brezis says "and thus $|a|^q \in L^{ (p/q)' } $", he is invoking Riesz's representation theorem: since the map $v \mapsto \int |a|^q v$ is a continuous linear functional on $L^{p/q}(\Omega)$, it follows that there exists a unique function $w \in L^{(p/q)'}(\Omega)$ such that $\int |a|^q v = \int wv$ $\forall v \in L^{p/q}$, from where, supposedly, it would follow that $w = |a|^q$, due to unicity.
The problem is that Riesz's theorem only guarantees the unicity in $L^{ (p/q)' }$; that is, it only tells us that there can't be two distinct functions both in $L^{ (p/q)' }$ representing the same continuous linear functional. Hence, we cannot directly apply (the unicity part of) such theorem to $|a|^q$ and $w$, since we have not yet proved $|a|^q \in L^{ (p/q)' }$ (in fact, that's what we would like to conclude!)
My question is, then: have I misinterpreted Brezis' argument or is it flawed? (Maybe he had something else in mind?)
 A: In other words, you have to prove the following:

Suppose $a$ is a measurable function such that, for all $v\in L^p$, $\int_\Omega av=0$. Then $a=0$ a.e.

This is immediate from the du Bois-Reymond Lemma. If you do not have this result (or you want to see this in general measurable spaces ($\sigma$-finite, perhaps)), you can use the characteristic functions of sets of the form $\left\{x:|v(x)|\geq 1/n\right\}$. Can you see how this solves your problem?
A: This first paragraph is not properly an answer but it may clarify things for future readers. The step from (S1) to
$$\int |a|^q |v| \leq C^q \| v \|_{p/q} \quad \forall v \in L^{p/q}(\Omega)$$
was to me a bit difficult to see but it is simply the fact that $v\in L^{p/q} \iff v^{1/q}\in L^{p}$. Raise (S1) to the power $q$ and apply it to $u = v^{1/q}$.
Let $\phi$ stand for $L^{p/q}(\Omega) \to \mathbb{K}, v \mapsto \int |a|^q v$. (Here $\mathbb{K}$ stands for $\mathbb{R}$ or $\mathbb{C}$.) This  $\phi$ is in $\left(L^{p/q}(\Omega)\right)^{*}$. Riesz Representation Theorem guarantees there is a unique $b$ in $L^{(p/q)^*}(\Omega)$ such that $\phi(v) = \int b v$ for all $v$ in $L^{p/q}(\Omega)$. Using this, the definition of $\phi$ and the linearity of integration, we get:
$$ \quad \int (|a|^q-b)v = 0 \quad \forall v \in L^{p/q}(\Omega).$$
So, as Luiz Cordeiro drew attention to, we want to show that this implies $(|a|^q-b) = 0$  almost everywhere.  It will follow from this:
$$
\left.
\begin{array}{l}
h: \Omega \to  \mathbb{K} \text{ is measurable}\\
\Omega \text{ is } \sigma\text{-finite}\\
\forall w\in L^r(\Omega) \;\; \int h w = 0
\end{array}
\right\} \implies h=0  \text{ a.e}.
$$
Proof. Since $\Omega$ is $\sigma$-finite, there are measurable subsets $X_n$ such that $|X_n|< \infty$ and $\bigcup_n X_n = \Omega$. Fix $n$ and let $E\subset X_n$ be measurable.
Since $|E| \leq |X_n|<\infty$, the indicator function $\mathbf{1}_{E}$ is in $L^{r}(\Omega)$ and so, by the property $h$ has, $\int h · \mathbf{1}_{E} = \int_{E} h  = 0$.
This is true for all measurable sets $E$ in $X_n$ and so $h = 0$ almost everywhere on $X_n$. Let us check this. Let $h_i = \mathrm{Im} h$ and $h_r = \mathrm{Re} h$ and write $h = i(h_i^+ -h_i^-)+ (h_r^+ -h_r^-)$ by decomposing $h_i$ and $h_r$ in negative and positive parts.  If $h$ were not $0$ almost everyhere on $X_n$, some $f$ in $\{h_i^+, h_i^-, h_r^+, h_r^-\}$ would have to be positive in a set of positive measure zero, and consequently the measure of $E = \{x\in X_n | f(x) >0\}$. If, say $f = h_i^+$, we would have $\mathrm{Im} \int_E h = \int_E h_i^+ > 0$ and then $\int_E h = i\mathrm{Im} \int_E h + \mathrm{Re} \int_E h$ would not be zero. The same would happen in the other cases. Therefore, $h$ is zero almost everywhere on $X_n$. In other words, $h=0$ in $X_n$ except perhaps on a null set $Z_n \subset X_n$. This can be said for all $n$ and so we conclude $h = 0$ almost everywhere on $\Omega$ by noticing that $h$ vanishes on $$\textstyle\bigcup_n (X_n \setminus Z_n) = \left(\bigcup_n X_n \right)\setminus \left(\bigcup_n Z_n\right) =  \Omega \setminus \left(\bigcup_n Z_n\right)$$
and that the set $\bigcup_n Z_n$ has measure zero. $\square$
