$g \in L^q(F)$ if for all $f \in L^p(F)$, $f(\cdot)g(\cdot)$ integrable? Here's a problem I'm struggling to prove.

Suppose $g(x)$ a measurable function on a measurable set $F$.
And for all $f \in L^p(F)\ (1\le p &lt; \infty)$, $f(\cdot)g(\cdot)$ is integrable. How can we prove that $g \in L^q(F)$ ? Here $\frac 1 p + \frac 1 q = 1$.

My thoughts are :

Let $g_n(x) = g(x)$ for all $x$ where $\vert g(x) \vert \le n$. Then we can show that $g_n(x) \in L^q(F)$, $g_n$ corresponds to a functional on $L_p(F)$, say $F_n$ , and $\Vert F_n \Vert$ is exactly $\Vert g_n \Vert_{L^q}$. 

With  $F_n(f) = \int f\cdot g_n\ d\mu \rightarrow \int f\cdot g &lt; \infty$ for all $f \in L^p(F)$, the Uniform-Bounded-Principle tells us that $F_n$ is uniformly bounded(and also $\Vert g_n \Vert$).

Here is the question in my proof. Can we conclude that $\int \vert g\vert ^q d\mu = \lim \int \vert g_n \vert ^q d\mu &lt; \infty$ ? (I don't think we can use the Dominated Convergence Theorem here since we don't know yet if $\vert g \vert ^q$ is integrable. 

(If the proof above is totally in the wrong direction, please feel free to point out.)
This is not true in general, we need some additional assumption like $\sigma$-finiteness for example.

For a counterexample in the general case consider $F = \{0,1\}$ equipped with the $\sigma$-algebra $\mathcal P(F)$ and a measure $\mu$ with $\mu(0) = 0, \mu(1) = \infty$. If we define $g = \chi_{\{1\}}$, where $\chi_A$ denotes the indicator function of $A$, then clearly $g\notin L^q$ if $q&lt;\infty$.
However, if $f\in L^p$, then $f(1) = 0$ and therefore $fg = 0$ is integrable.

Your attempted proof for the case $p&gt;1$ is almost good, it just needs a slight refinement.

If we consider $\sigma$-finiteness (although a weaker assumption might be enough aswell) we find a sequence $E_n$ of measurable sets with finite measure and $E_n\nearrow F$. Let us further assume that $g\geq 0$.1 Let now $g_n := \min\{n, g\}\cdot \chi_{E_n}$. For these $g_n$ it is clear that $g_n\in L^q$, since $$\int |g_n|^q \leq \int_{E_n} n^q = n^q\mu(E_n)&lt;\infty.$$
For the ones defined in your proof this is not so clear. Now you conclude that $\int |g_n|^q$ are uniformly bounded as in your question. But also $g_n\to g$ monotonously(!), so the theorem of monotone convergence yields $\int|g_n|^q\to \int|g|^q$.

For the case $p=1$ let $T:L^1\to \mathbb C, f\mapsto \int fg$.
The uniform boundedness principle yields $\|T\|&lt;\infty$ with the same proof as in the case $p&gt;1$. Assume that $\mu\{g&gt;\|T\|\} &gt; 0$. By $\sigma-$finiteness, we find $A\subset \{g&gt;\|T\|\}$ with $0&lt;\mu(A)&lt;\infty$.
But now set $f := \frac{1}{\mu(A)}\chi_A\in L^1$. Then $\|f\|_1 = 1$ and we conclude

$$\|T\| = \frac{1}{\mu(A)}\int \|T\|\chi_A &lt; \frac{1}{\mu(A)}\int g\chi_A = \int gf \leq \|T\|\|f\| = \|T\|,$$ which is a contradiction and therefore shows $\|g\|_\infty \leq \|T\|&lt;\infty$.

1Otherwise we pass to $|g|$ instead of $g$.
Since $L(f) = \int gf$ is a functional on $L^p$ by hypothesis, it must be that $L(f) = \int hf$ for some $h\in (L^p)^* = L^{q}$ where $p^{-1} + q^{-1} = 1$ by the Riesz representation theorem. Then you have $\int (g-h)f = 0$ for all $f\in L^p$, implying $g = h$ almost-everywhere.