I am looking at the proof of the Lebesgue decomposition theorem. Let $\mu,\nu$ be finite measures.
Here, they first define a measure $\nu^o$ which is absolutely continuous with respect to $\mu$. Now they set $\nu^\perp := \nu - \nu^o$, and show that $\nu^o$ is maximal among all measures $\rho$ such that $\rho \le \nu$ and $\rho \ll \mu$.
Using this they show Orthogonality. Let $\tau$ be a measure such that $\tau \le \mu$ and $\tau \le \nu^\perp$. Clearly, this implies that $\nu^o + \tau \le \nu$ and $\nu^o + \tau \ll \mu$. By the maximality, $\nu^o + \tau \le \nu^o$ and we conclude that $\tau = 0$ and $\nu^\perp \perp \mu$.
I don't follow the final line. How does $\tau=0$ conclude that $\nu^\perp \perp \mu$?
I suppose $\mu$ and $\nu$ positive measures. Let $f$ be the Radon-Nikodym derivative of $\mu$ w.r.t. $\mu+\nu^{\perp}$ and $g$ be the Radon-Nikodym derivative of $\nu^{\perp}$ w.r.t. $\mu+\nu^{\perp}$. Let $\tau (E)=\int_E \min \{f,g\} d (\mu+\nu^{\perp})$. Then $\tau \leq \mu$ and $\tau \leq \nu^{\perp}$ so $\tau =0$. Hence $\int_E \min \{f,g\}d(\mu+\nu^{\perp})=0$ for every measurable set $E$ which implies $\min \{f,g\}=0$ almost everywhere w.r.t. $\mu+\nu^{\perp}$ (hence w.r.t. each of $\mu$ and $\nu^{\perp})$. This means one of $f,g$ is $0$ at every point excluding a null set w.r.t. each of $\mu$ and $\nu^{\perp}$. Can you now prove that $\mu \perp \nu^{\perp}$?