Here I state the proposition:
Let $(X, \mathcal{F}, \mu)$ be a measure space and $f: X \to \mathbf{R}$ be a $\mu$-semi-integrable function. Then there exists a signed measure $\lambda: \mathcal{F} \to \overline{\mathbf{R}}$ such that for all $E \in \mathcal{F}$, we have $$ \lambda(E) = \int_E f \,d\mu. $$ Moreover, we have $\lambda: \mathcal{F} \to \overline{\mathbf{R}}$ satisfies the property that if $A \in \mathcal{F}$ is such that $\mu(A) = 0$, then $\lambda(A) = 0$. Furthermore, if $g: X \to \mathbf{R}$ is a $\lambda$-semi-integrable function, then $gf: X \to \mathbf{R}$ is a $\mu$-semi-integrable function and $$ \int_X g \,d\lambda = \int_X g f \,d\mu. $$
I am having trouble to show the bolded part of the statement. In particular, why is it true that $gf$ is a $\mu$-simi-integrable function? In fact, I am having doubts on if this is even true. My attempts so far are considering the expansion of $(f + g)^2$ and extract $fg$ from it and analyze the integrability of each of the terms in the expansion. However, I do not see how this would work out as $f^2$ and $g^2$ are not even necessarily semi-integrable even if $f$ and $g$ are. If we have integrability in the usual sense (both the positive part and negative part are finite), then we could have used Hölder. This doesn't seem to work here? I am not sure what I am missing here.
Definition of Semi-Integrable: We say a measurable function is semi-integrable if either the positive or the negative part of the function integral is finite.
Update: I was able to show the result for the case where $g$ is a nonnegative simple function. However, I was not able to generalize the result to the nonnegative measurable function case using monotone convergence. In particular, I am stuck on proving the product $gf$ is $\mu$-semi-integrable without any assumptions on the boundedness of $g$, given that $g$ is a nonnegative measurable function and hence automatically $\lambda$-semi-integrable. I am not sure how to tackle this problem.
The product of two semi-integrable functions need not be semi-integrable. Assume that $\mu$ is finite and there exists $A\in\mathcal F$ and a positive measurable function $h$ with $\int_Ahd\mu=\int_{X\setminus A} h d\mu =\infty$. Denoting by $1_A$ the indicator function (with values $1$ on $A$ and $0$ on $X\setminus A$, you get two semi-integrable functions $f=1_A+h1_{X\setminus A}$ and $g=-h1_A +1_{X\setminus A}$ such that the product $fg=-h1_A +h1_{X\setminus A}$ is not semi-integrable.
For a concrete example you can take the Lebesgue measure on $[0,1]$ and $A= [0,1/2]$ or, even more elemntary, any finite measure on $\mathbb N$ with $\mu(\{n\})>0$ for all $n\in\mathbb N$ and $A$ the even integres.
Edit. Your comment made me realize that this does not answer your question. Since $\lambda$ is only a signed measure one has to be careful with the definition of the integral (which is based very much on positivity and monotonicity). The simplest way here is to decompose $\lambda$ as the difference of two positive measures $\lambda^+-\lambda^-$ which <have the $\mu$-densities $f^+=\max\{f,0\}$ and $f^-=\max\{-f,0\}$. Writing a measurable function $g$ as $g=g^+-g^-$ I would call $g$ semi-integrable with respect to $\lambda$ if the expression $$\int g d\lambda=\left(\int g^+ d\lambda^+ +\int g^- d\lambda^-\right) - \left(\int g^+d\lambda^- +\int g^- d\lambda^+\right)$$ is well-defined (i.e., not of the form $\infty-\infty$). Then the reduction of the formula $\int g d(f\mu)=\int gfd\mu$ to the case of positive functions should be a tedious but routine checking of cases.