What should I use in order to get the required function $g \in L^1$? Let $f \in L^1$ be such that $f$ is not equivalent to any bounded function. Prove there exists a function $g \in L^1$ such that $fg \notin L^1$.
I know that $m( \{x:|f(x)|>M\} )>0$, $\forall M>0$, but I can't see to come up with anything else useful to use for this problem in order to get the required function. 
Thank you for any hints as to how may I go about approaching this problem.
 A: I don't think there is a simple measure theoretic proof of this. Completeness of $L^{1}$ is crucial. Here is the standard functional analytic proof: define linear functionals $\Phi_n$ on $L^{1}$ by $\Phi_n (g)=\int I_{|f| \leq n} fg$. These are continuous linear functionals on $L^{1}$ and $\sup_{\{n \geq 1\}} |\Phi_n (g)| <\infty$. Now apply Uniform Boundedness Principle to conclude that $\sup ||\Phi_n|| <\infty$. A standard argument used in the proof of $(L^{1})^{*} =L^{\infty}$ show that $||\Phi_n||=||I_{|f| \leq n} f||_{\infty}$. Hence $f$ is (essentially) bounded. 
A: One can give a simple constructive proof of existence of such $g$.
Assume we have $f\in L^1(0,1)$ which is not $L^\infty$ (we fix $(0,1)$ for the sake of clarity, this does not impose any restriction on the construction below). For each $n\in \mathbb{N}$ define 
$$
E_n = \{ x \in (0,1): n\leq |f(x)|  \}.
$$
Since $f\notin L^\infty$, then $\mu(E_n) >0$ for all $n\in \mathbb{N}$.
The aim now is to construct a function $g\in L^1$ such that $fg \notin L^1$.
To this end, set
$$
(1) \qquad g = \sum_{n=1}^\infty \frac 1n a_n \chi_{E_n},
$$
where $a_n>0$ is a sequence of positive numbers to be fixed in a moment. 
We thus have
$$
(2) \qquad \int|fg| d\mu =  \sum_{n=1}^\infty \frac{a_n}{n} \int_{E_n}|f| d\mu \geq \sum_{n=1}^\infty \frac{a_n}{n}  n \mu(E_n) = \sum_{n=1}^\infty a_n\mu(E_n).
$$
Hence, to complete the construction we need to fix $a_n$ so that
(a) $\sum_{n=1}^\infty  \frac{1}{n} a_n \mu (E_n)  < \infty$ which is the condtion of $g\in L^1$ (see $(1)$ )
(b) $\sum_{n=1}^\infty a_n \mu (E_n) = \infty$, which amounts to $fg \notin L^1$
(see $(2)$ ).
Thanks to $f\notin L^\infty$, we have $\mu(E_n) >0 $ for all $n$ and hence
setting $a_n = \frac{1}{\mu(E_n)} \frac{1}{n^{1/2}}$ we get both (a) and (b)  and hence complete the construction of $g$.
