Given a sigma algebra $\mathcal{F}$, is there a natural topology worth defining on it? More specifically, is there a topology you can put on $\mathcal{F}$ which ensures a measure of interest $\mu: \mathcal{F} \to \mathbb{R}$ is continuous? Moreover are there any known guarantees regarding whether the resulting topological space is triangulable?
EDIT: Thanks for the answers everyone. I thought I'd explain a little bit more why I am interested in this problem. In Topological Data Analysis it is fairly common to investigate the persistent homology of the sublevel sets of density functions. I wanted to know if something similar could be done for probability measures. That is, is it possible to compute persistence diagrams for probability distribution functions. To do so, I need an underlying topology for $\mathcal{F}$ for which 1) the topology is triangulable and 2) any measure $\mu$ on $\mathcal{F}$ is tame. Now, I don't plan to actually compute these persistent diagrams, but I wondered if there would be any provable relationship between the persistence diagram of a distribution function and the persistence diagram of its corresponding density. To do so, I first need to verify whether computing a persistence diagram on the sublevel sets of a distribution function even "makes sense".
EDIT 2: I am using the following definition of tame functions, from here:
This is more of a comment, but far too long. From the point of view of measure theory, there is no natural T2 topology on the $\sigma$-algebra of a measure space~$(X,\sigma,\mu)$. Compare the case of subsets of a metric space $(Y,d)$. In order to obtain a reasonable topology, one needs not distinguish between $A$ and $\overline A$ (closure), which are indeed indistinguishable from the point of view of the functional $d(A,\cdot)$ (the analogue of the functional $\mu$ in this example). In this case one restricts one's attention to closed subsets, defining the Hausdorff topology on the set of closed subsets of $(X,d)$.
For measure-theoretical purposes (but I am sure the OP's interest is beyond this, because of the triangulability question), the natural thing to do is identify two sets $A_1,A_2\in \Sigma$ if $\mu(A_1\triangle A_2)=0$ ($\triangle$ being the symmetric difference). More formally, let $\Sigma^\mu$ be the Carathéodory $\mu$-completion of $\Sigma$ and $\mathcal N^\mu$ be the $\sigma$-ideal of $\mu$-negligible subsets of $X$. The measure algebra $(\mathfrak A, \bar\mu)$ of $(X,\Sigma,\mu)$ is defined as the quotient of $\mathfrak A$ of $\Sigma^\mu$ by $\mathcal N^\mu$, endowed with the quotient functional $\bar\mu$. In the following, write $a=[A]\in\mathfrak A$ for the class of $A\in\Sigma$, and note that $0=0_{\mathfrak A}=[\varnothing]$ and $1=1_{\mathfrak A}=[X]$ (as usual the product is just $\cap$, and $\mathfrak A$ is a Bolean algebra with this product).
At this point I should remark that much information on both the space $X$ and the $\sigma$-algebra $\Sigma$ is lost, since there is a Procrustean bed of measure spaces with isomorphic measure algebras.
For simplicity, I assume from now on that $\mu$ is finite. Much of what follows can be generalized to the $\sigma$-finite case at least. A natural uniformity is defined on $(\mathfrak A,\bar \mu)$ by the family of pseudo-metrics (triangle inequality, symmetry, but possibly vanishing outside the diagonal): $$\rho_a(b,c):= \bar\mu\big(a\cap (b\triangle c)\big)\,.$$
The topology induced by this uniformity is very closely related to the topology of convergence in measure on the space of functions $L^0(\mu)$. When $\bar \mu$ is finite (here: by assumption), the uniformity (hence the topology) is induced by the
pseudo-metric $\rho_1$. From the point of view of measure theory, this topology is quite natural, e.g., since it allows to neglect the fine properties of adherence of $\Sigma$ to $X$.