Set of finite signed measures with the weak*-topology is a topological vector space For a compact space $X$ I want to show that the space of finite signed measures  on the Borel-$\sigma$-algebra on $X$, equipped with the weak*-topology, i.e. the corsest topology such that
$\mu\mapsto\int f~\mathrm{d}\mu~$ is continuous for all $f\in C_b(X)$ is a topological vector space.
Therefore I think I need to show that the maps $(\mu,\nu)\mapsto\mu+\nu$ and $(c,\mu)\mapsto c\mu$ are continuous.
I tried to do it with the formal definition of continuity in topological spaces, i.e. I wanted to show that $\{(\mu,\nu):\mu+\nu\in U\}$ and $\{(c,\mu):c\mu\in U\}$ are open in the topology for open $U$ but I have no idea how to do that.
Can someone give me a reference or a hint on how to do that?
 A: Perhaps you should do it by nets:
Assume that $(\mu_{\alpha},\nu_{\alpha})\rightarrow(\mu,\nu)$ weak$^{\ast}$, then $\displaystyle\int fd(\mu_{\alpha}+\nu_{\alpha})=\int fd\mu_{\alpha}+\int fd\nu_{\alpha}\rightarrow\int fd\mu+\int fd\nu=\int fd(\mu+\nu)$, so $\mu_{\alpha}+\nu_{\alpha}\rightarrow\mu+\nu$ weak$^{\ast}$. For the scalar multiplication is similar.
A: There are a lot of details here which I hope I did not get confused, but here is my attempt:
Denote: 
$$T:\mathbb{R}\times \mathcal{M}\rightarrow \mathcal{M}, \quad T(c,\mu)=c\cdot \mu \qquad \text{and} \qquad S:\mathcal{M}\times \mathcal{M}\rightarrow \mathcal{M}, \quad S(\mu,\nu)=\mu+\nu$$
An element in the sub-basis of the weak-* topology is of the form:
$$ U_{\epsilon,f}(\mu):= \Big\{ \nu \in \mathcal{M} \Big\vert \; \vert \nu(f)-\mu(f)\vert<\epsilon  \Big\} \quad \text{where} \quad \epsilon>0 \; \text{and} \; f\in C_b(\mathbb{R}) $$
For the sake of simplicity, deal with neighbourhoods of the form $U_{\epsilon,f}(0)$ and that $\epsilon$. Let $(c,\nu)\in T^{-1}\Big[ U_{\epsilon,f}(0) \Big]$ and denote $\delta:=\vert \nu(f)\vert$. Then:
$$ \vert c'\eta(f) \vert  \leq \vert c\nu(f) \vert+ \vert c\nu(f)-c'\eta(f)\vert  $$
$$ \leq  \vert c\nu(f) \vert + \vert (c-c')\nu(f) \vert+\vert c'\nu(f)-c'\eta(f)\vert \leq $$
$$ \leq \vert c\nu(f)\vert+\vert c-c'\vert \cdot \vert \nu(f) \vert + \vert c'\vert \cdot \vert \nu(f)-\eta(f) \vert $$
If $c\cdot\delta=0$, then by the above bound, every  $(c',\eta)\in(-\epsilon,\epsilon)\times U_{\epsilon,f}(\nu)$ is contained in $T^{-1}\Big[ U_{\epsilon,f}(0) \Big]$, and $(c,\nu)$ is an interior point of $T^{-1}\Big[ U_{\epsilon,f}(0) \Big]$. Else $\Big( c-\frac{\epsilon-c\delta}{2\delta}, c+\frac{\epsilon-c\delta}{2\delta} \Big)\cap[ -c/2,c/2  ]^c\times U_{\frac{\epsilon-c\delta}{2},f}(\nu)$ is an open neighbourhood of $(c,\nu)$ contained in $T^{-1}\Big[ U_{\epsilon,f}(0) \Big]$. In both cases, it shows that $(c,\nu)$ is an interior point of the pre-image, and hence $T^{-1}\Big[ U_{\epsilon,f}(0) \Big]$ is open. This 'roughly' shows that $T$ is continuous, up to choice of the center of the basic neighbourhoods. For $(\nu,\mu)\in S^{-1}\Big[ U_{\epsilon,f}(0) \Big]$, we can find a bound:
$$ \vert \eta_1(f)+\eta_2(f)\vert \leq \vert \mu(f)+\nu(f)\vert+\vert \mu(f)-\eta_1(f)\vert+ \vert \nu(f)-\eta_2(f)\vert $$
And denoting $\delta':=\vert \eta_1(f)+\eta_2(f)\vert$, we can conclude that  $U_{(\epsilon-\delta)/2,f}(\nu) \times U_{(\epsilon-\delta)/2,f}(\mu)$ is an open neighbourhood of $(\mu,\nu)$ contained in $S^{-1}\Big[ U_{\epsilon,f}(0) \Big]$, and we can conclude that $S$ is continuous.
It still remains to explain why it is enough to work with $U_{\epsilon,f}(0)$ and not general $U_{\epsilon,f}(\mu)$. This is more than a hint, because I struggled with formalizing it in my head.
