Poisson Process independent Wiener Process using singular measures I was reading some stochastic calculus of Jump processes and saw the following result:

If $W_t$ is Brownian and $N_t$ is Poisson both adapted to the $W_t$'s natural filtration then these processes are independent.  

The book went on to give a stochastic calculus argument.
I was curious couldn't we just prove this by using the fact that the increments of $W_t$ is absolutely continuous wrt Lebesgue measure $m$ and the increments of $N_t$ are absolutely continuous wrt the counting measure $c$ which is singular wrt $m$....
This is just an intuition but I do not know how to approach the proof.  So I'm asking here...
 A: No, the singularity of their laws has nothing to do with the singularity of the Lebesgue and counting measures. 
There are many ways to show this singularity. The simplest is to say that the paths of $N$ live in $\mathbb{Z}^{[0,+\infty)}$, while those of $W$ belong here with probability $0$, whence the required singularity follows.

Now concerning the independence. Though the links given in the comments are relevant, there is some confusion. In particular, I believe that the accepted answer of @kakaritsu here is wrong. Below I will comment on this. 
First of all, the statement is absurd as it is written now. The Poisson process cannot be at the same time adapted to the natural filtration of $W_t$ and be independent of it. No way.
Let us try to formulate the true statement. The obvious alternative is

Let $(\Omega, \mathcal F, \mathbb{F}=\{\mathcal F_t\},\mathsf P)$ be a filtered probability space, and let the processes $W$ (Wiener), $N$ (Poisson) be $\mathbb F$-adapted. Then they are independent.

Still wrong. The problem with this formulation is that any couple of processes is adapted to some filtration. Ok, let us try something better. Namely, call a process $W$  $\mathbb F$-Wiener if 1) it is a Wiener process; 2) it is $\mathbb F$-adapted; 3) for each $t>0$, the increments of $W$ after $t$ are independent of $\mathcal F_t$. A $\mathbb F$-Poisson process is defined similarly. Now a reasonable alternative is: 

Let $(\Omega, \mathcal F, \mathbb{F}=\{\mathcal F_t\},\mathsf P)$ be a filtered probability space, and let the processes $W$, $N$ be $\mathbb F$-Wiener and $\mathbb F$-Poisson. Then they are independent.

It is hard to believe, but this is still wrong. The reason is following: despite $W_s-W_t$ and $N_s-N_t$ are independent of $\mathcal F_t$ for $s\ge t$, the pair $(W_s-W_t,N_s-N_t)$ can depend on it! So the true formulation is:

Let $(\Omega, \mathcal F, \mathbb{F}=\{\mathcal F_t\},\mathsf P)$ be a filtered probability space, and let $W$, $N$ be $\mathbb F$-adapted Wiener and Poisson processes such that $(W_s-W_t,N_s-N_t)$ is independent from $\mathcal F_t$ for each $0<t<s$. Then $W$ and $N$ are independent.

Alternatively, one can forget about the stochastic basis and formulate it as follows: 

Let the processes $W$, $N$ be Wiener and Poisson such that $(W,N)$ has independent increments. Then $W$ and $N$ are independent.

The both last statements are correct, see Kallenberg Lemma 13.6 for the proof.
Now why I think the above cited answer is incorrect (and it was criticized for this by @Did). In fact, the answerer refers to the Ito formula (nothing is wrong here) and then claims that $L$ is a martingale. And the last is wrong in general. For that one need some extra assumptions like those in the last two formulations. 
I may be wrong somewhere, so any comments (@Did, if you please?) are welcome.
