Prove that (sigmaleft(M_{s}-W_{s}, s leq t ight)=sigmaleft(W_{s}, s leq t ight)). This equality follows from (int_{0}^{t} mathbb{1}_{left{M_{s}-W_{s}=0

Prove that \(\sigma\left(M_{s}-W_{s}, s \leq t\right)=\sigma\left(W_{s}, s \leq t\right)\).

This equality follows from \(\int_{0}^{t} \mathbb{1}_{\left\{M_{s}-W_{s}=0\right\}} d\left(M_{s}-W_{s}\right)=M_{t}\). Use the fact that \(d M_{s}\) is carried by \(\left\{s: M_{s}=B_{s}\right\}\).