$f\in L_1(E)$ be not equivalent to any bounded function. Show that $\exists g\in L_1(E) $ such that $f.g\notin L_1(E)$. Let $f\in L_1(E)$ be not equivalent to any bounded function. Show that $\exists g\in L_1(E) $ such that $f.g\notin L_1(E)$.
I don't know how to approach to this question. Any help is appreciated.
 A: First note that this is wrong (depending on what you mean by equivalence) if the measure is not at least assumed to be semifinite. For a counterexample consider $E = \Bbb{N}$ with the power set as $\sigma$-algebra and $\mu(\{2n\}) = \infty$ for each $n \in \Bbb{N}$ (the "rest" of the measure can be chosen arbitrarily). Then $f = \sum_n n \cdot \chi_{\{2n\}}$ is not equivalent (i.e. equal a.e.) to a bounded function, but $f\cdot g \in L^1(\mu)$ for all $g \in L^1(\mu)$.
So let us assume that $\mu$ (or $E$) is semifinite.
Assume that the claim is wrong. This implies that the linear operator
$$T : L^1(E) \rightarrow L^1(E), g \mapsto f\cdot g$$
is well-defined.
For $g_n \rightarrow g$ in $L^1(E)$ and $Tg_n \rightarrow h$ in $L^1(E)$, we get $g_{n_k}(x) \rightarrow g(x)$ a.e. and $f(x) \cdot g_{n_{k_{\ell}}}(x) = Tg_{n_{k_{\ell}}}(x) \rightarrow h(x)$ a.e. for suitable subsequences.
Using this, we conclude $h(x) = f(x) \cdot g(x)$ a.e., which means $h = Tg$. Using the closed graph theorem, we conclude that $T$ is bounded.
By assumption, the set $E_n := \{x \in E \mid |f(x)| > n\}$ has positive (possibly infinite) measure. We are assuming that $E$ is semifinite, which gives us a set $F_n \subset E_n$ of positive, finite(!) measure.
Now $$n \cdot \mu(F_n) \leq \int_{F_n} |f(x)| d\mu =\Vert T(F_n) \Vert \leq \Vert T \Vert \cdot \Vert F_n \Vert = \Vert T \Vert \cdot \mu(F_n),$$
i.e. $\Vert T \Vert \geq n$ for all $n \in \Bbb{N}$, contradiction.
