I guess you can show that total variation is a measurable function, but not sure whether it will help to answer which Borel class does $B$ belong to.

Not sure whether this answers your question, but Lemma 2 of "On the existence of good stationary strategies" by Sudderth may be of use.

Density is just one representation of a probability measure (through another measure) and it is useful as a tool, when it comes up naturally, not as a goal: by itself it does not tell anything new about probabilities. Now, in the space you've described a natural measure would be the Wiener measure as @Jochen mentioned, but if I'm not mistaken your $P_X$ is singular w.r.t. the Wiener measure. The question is what were you going to do after you'd get a density - if you specify that, I guess the conclusion would be that you don't really need density.

@MichaelGreinecker: certainly depends on the tone. you should have taken my original comment with a grain of salty joke :) I'm going to ask a question regarding the history of probability and measure theory on MSE in 5 minutes, may be of your interest.

@MichaelGreinecker: oh no, have you just critisized Kolmogorov?

@MichaelFanZhang: it may become unbounded ($-\infty$), but that's in fact important to negative dynamic programming. Also, since $g$ is sign-semidefinite, the sum is always negative so its expectation is well-defined, even though it may be $-\infty$ as well. Did I answer your question?

@Burak: I see your point, indeed some Borel selection would be needed.

I am also aware of the updated version of Wagner's paper published in 1979, where he claims to add more results know in the Russian literature by that time, however I do not have an access to that.

@Burak: for some reason I was not notified, so I just came across your comment. Do you mean here, that if $E$ is smooth then $X/E$ is at least analytic, and if $X/E$ is a standard Borel space, then $E$ is smooth? Also, isn't the surjectivity in the 2nd sense (meeting each $F$ class) equivalent to bireducibility by definition?

I was exactly trying to use a Borel set which does not have a Borel uniformization, and relate it to a $\mathrm{Gr}(f)^{−1}$ via an isomorphism, however that did not work. Your idea of using a projection is really nice, thanks. Also, there is a survey of measurable selection theorems (Wagner 1977-79). Do you know if there was any progress after that, maybe another survey paper? Please tell me in case I should make it a separate question.

@JoelDavidHamkins: my hypothesis that existence of a surjective $g$ implies existence of a Borel selector is not correct. If such a surjection $g$ exists, and there is a Borel selector $h:X \to X$ for $\sim_g$, then $h$ is $g$-measurable, and hence it factors through $g$ as $h = s\circ g$ where $s:Z\to X$ is Borel. The fact that $s\circ g$ is a Borel selector for $\sim_g$ is equivalent to $g\circ s = \mathrm{id}_Z$, and according to your answer such $s$ may fail to exist.