This feels like a silly question. I'm reading somewhere that
$$ \mathbb{P} (Y = 1,h(x)=-1\mid X=x) = \mathbb{P}(Y=1\mid X=x) \mathbb{ I }_{\{h(x) = -1\}} $$
where $\mathbb{ I }$ is the indicator function (not sure the best way to LaTeX this). Here $h$ is a deterministic classifier. I'm wondering why we can just 'do' this, it doesn't feel like this aligns with my common sense of manipulating probability equations. ie. Can we treat the events $Y=1$ and $h(x) = -1$ conditionally independent given $X=x$?
This come frome the fact that if $X$ is $\mathcal G-$measurable, then $$\mathbb E[XY\mid \mathcal G]=X\mathbb E[Y\mid \mathcal G].$$
Since you have that $\boldsymbol 1_{\{h(X)=-1\}}$ is $\sigma (X)-$measurable, you have that \begin{align*} \mathbb P\{Y=1, h(X)=-1\mid X\}&:=\mathbb E[\boldsymbol 1_{\{Y=1\}}\boldsymbol 1_{\{h(X)=-1\}}\mid X]\\&=\boldsymbol 1_{\{h(X)=-1\}}\mathbb E[\boldsymbol 1_{\{Y=1\}}\mid X]\\ &=\boldsymbol 1_{\{h(X)=-1\}}\mathbb P\{Y=1\mid X\}. \end{align*}