Rudin RCA Problem 6.4 I'm trying to solve the following Problem 6.4 from Rudin's RCA:
Suppose that $1\leq p\leq \infty$, and $q$ is the exponent conjugate to $p$. Suppose that $\mu$ is a $\sigma$-finite measure and $g$ is a measurable function such that $fg\in L^1(\mu)$ for every $f\in L^p(\mu)$. Prove that then $g\in L^q(\mu)$.
I can get to the point where $\int_{E}|g|d\mu<\infty$ for every finite measure set $E$, but I'm not sure where to proceed. I would appreciate any suggestions.
Thanks!
 A: Here are a few hints: you can define $$X_k = \bigg\{ f \in L^p(\mu) : \|fg\|_1 \le k \|f\|_p \bigg\}.$$ By hypothesis $\cup_{k=1}^\infty X_k = L^p(\mu)$.
Show that each $X_k$ is closed. Then, according to the Baire Category Theorem, at least one $X_k$ has a nonempty interior. This means there exist $k \in \mathbf N$,  $f_0 \in L^p(\mu)$, and $\epsilon > 0$ with the property that if $f \in L^p(\mu)$ and $\|f - f_0\| \le \epsilon$, then $f \in X_k$.
If $f \in L^p(\mu)$ and $\|f\|_p \le 1$, then $\|(\epsilon f + f_0) - f_0\|_p \le \epsilon$, so that $\|(\epsilon f + f_0) g\|_1 \le k \|\epsilon f + f_0\|_p$. Use this fact to find a constant $C$ (depending on $\epsilon$, $k$, and $f_0$) such that
$$\sup_{\|f\|_p \le 1} \|fg\|_1 \le C.$$
Since the underlying space is $\sigma$-finite this gives you what you need.
A: Here is an argument using Uniform Boundenes Principle: Let $\mu(A_n) <\infty$ and $A_n$ increase to the whole space. Define $T_nf=fg\chi_{A_n \cap \{x: |g(x)| \leq n\}}$. Verify that $T_n$ is a bounded linear functional and $\|T_n\|^{q}=\int_{{A_n \cap \{x: |g(x)| \leq n\}}}|g|^{q} d\mu$. By hypothesis $\sup_n \|T_nf\|<\infty$ for each $f \in L^{p}$. It follows that $\sup_n \|T_n||<\infty$ But this is precisely the satatment that $g \in L^{q}$.
[For computation of $\|T_n\|$ just follow the argument in $(L^{p})^{*}=L^{q}$].
