Let $(X, \mathcal{S}, \mu)$ be a measure space and let $M(\mathcal{S})$ be the vector space of all complex measures on $(X,\mathcal{S})$. The total variation norm $\Vert \nu \Vert := |\nu|(X) $ makes $M(\mathcal{S})$ into a Banach space.
I want to show that $$\{h d \mu: h \in \mathcal{L}^1(\mu)\}$$ is a closed subspace of $M(\mathcal{S})$. Here $h d \mu$ is the complex measure $$A \mapsto \int_A h d \mu$$
Attempt: I managed to show it is a linear subspace, but I'm stuck at showing it is closed.
I tried the following:
Let $\nu_{n}:= h_n d \mu$ be a sequence with $\Vert\nu_n- \nu \Vert \to 0$ when $n \to \infty$. We want to show that there is $h \in \mathcal{L}^1(\mu)$ with $\nu = h d \mu$.
However, I'm not sure how to construct this $h$. A guess would be to try $h:= \lim_n h_n$ or even better $h:= \liminf _n h_n$ because then I don't need to show that the limit exists. But I don't know why this function would be integrable.
Some useful information: we have $$\nu(A) = \lim_n\int_A h_n d \mu$$ for all measurable subsets $A$. How can I construct $h$?
A hint suffices! I don't want a complete solution.
If $\nu_n =h_n d\mu$ converges in total variation then $\|\nu_n-\nu_m\|=\int |h_n-h_m|d\mu \to 0$. Use completeness of $L^{1}(\mu)$ to finish.