There's a step I'm not following in the proof of Girsanov theorem.
I've looked in "Brownian Motion, Martingales, and Stochastic Calculus" by Le Gall (Theorem 5.22) and in "Foundations of Modern Probability" by Kallenberg (Theorem 18.19) - and they both do pretty much the same thing.
Reminder:
Suppose $\Sigma=(\Sigma_t)_{t\ge 0}$ is a complete and right-continuous filtration, and $\mu\cong\nu$ are probability measures on $\Sigma_{\infty}$. Denote $D_t:=\frac{d\nu}{d\mu}\big\vert_{\Sigma_t}$. If $(M_t)_{t\ge 0}$ is a continuous local martingale under $\mu$, then $\hat{M}_t=M_t-[M,\mathcal{E}\left(D\right)]_t$ is a continuous local martingale under $\nu$.
Where I've denoted by $\mathcal{E}$ the stochastic exponentiation: $\mathcal{E}(X):=\exp\left(X_t-\frac{1}{2}[X_t]\right)$), and by $[X,Y]$ "the bracket" (=covariation, $\approx$polarization of the quadratic variation).
The step that eludes me is (I think) the claim - $$[M,D]_t=[\hat{M},D]_t$$ that I guess is equivalent to - $$\big[[M,\mathcal{E}(D)],D\big]_t=0$$
Why does it hold? Is it really just an immediate consequence of the construction - $$\left[(V\cdot M),N\right]_t=\left(V\cdot\left[M,N\right]\right)_t$$ and Ito's lemma (where $(V\cdot M)_t:=\int_0^tV_sdM_s$)? How so?
Recall that if $M,N$ are continuous local martingales then $[M,N]$ is the unique finite variation process such that $MN - [M,N]$ is a continuous local martingale and we have that $$\sum_{\pi_n} (M_{t_i^n}- M_{t_{i-1}^n})(N_{t_i^n}- N_{t_{i-1}^n}) \to [M,N]_t$$ in probability as the mesh of the partition $\pi_n$ goes to $0$.
So it will be enough to show that if $V$ is a finite variation process then $[V,N] = 0$ for a continuous local martingale $N$. Write $\|V\|_{1,T} = \sup \sum |V_{t_i} - V_{t_{i-1}}|$ for the $1$-variation of $V$ where the $\sup$ is over all partitions of $[0,T]$. Then we have for a partition $\pi_n$ of $[0,t]$, $$\bigg | \sum_{\pi_n} (V_{t_i^n} - V_{t_{i-1}^n})(N_{t_i^n}- N_{t_{i-1}^n}) \bigg | \leq \|V\|_{1,t} \max_{\pi_n} |N_{t_i^n}- N_{t_{i-1}^n}|$$ and the right hand side goes to $0$ as the mesh of $\pi_n$ goes to $0$ since $N$ is uniformly continuous on $[0,t]$. This implies that $[V,N]_t = 0$ as desired.