# What's going on with this inequality?

Let $$(\mathsf{X}, \mathcal{X})$$ be a measurable space, $$(\mathcal{F}_b(\mathsf{X}, \mathcal{X}), \Vert \cdot \Vert_{\infty})$$ be the space of all bounded measurable functions on this space equipped with the $$\sup$$ norm, $$(\textrm{M}(\mathsf{X}, \mathcal{X}), \Vert \cdot \Vert_{TV})$$ be the space of all finite signed measures equipped with the total variation norm.

If $$\xi \in \textrm{M}(\mathsf{X}, \mathcal{X})$$ and $$K$$ is a transition kernel, why is it true that $$|\xi_{+}(\mathsf{X})Kf(x) - \xi_{-}(\mathsf{X})Kf(x')| \le \Vert \xi_{+}(\mathsf{X})K(x,\cdot) - \xi_{-}(\mathsf{X})K(x',\cdot) \Vert_{TV} \Vert f \Vert_{\infty}$$ for any $$x,x' \in \mathsf{X}$$?

Define $$\mu_{x,x'}(\cdot) = \xi_{+}(\mathsf{X})K(x,\cdot) - \xi_{-}(\mathsf{X})K(x',\cdot)$$ and let $$H_{x,x'}$$ be the Jordan set for it. Then \begin{align*} &|\xi_{+}(\mathsf{X})Kf(x) - \xi_{-}(\mathsf{X})Kf(x')| \\ &= |\mu_{x,x'}(f)| \\ &= |\mu_{x,x'}(f \mathbb{1}_{H_{x,x'}}) + \mu_{x,x'}(f \mathbb{1}_{H_{x,x'}^c})| \\ &\le |\mu_{x,x'}(f \mathbb{1}_{H_{x,x'}})| + |\mu_{x,x'}(f \mathbb{1}_{H_{x,x'}^c})| \\ &\le \Vert f \Vert_{\infty}\left[ |\mu_{x,x'}(H_{x,x'})| + |\mu_{x,x'}(H_{x,x'}^c)|\right]\\ &= \left(\mu_{x,x'}^+(\mathsf{X}) + \mu_{x,x'}^-(\mathsf{X})\right)\Vert f \Vert_{\infty} \\ &= \Vert \xi_{+}(\mathsf{X})K(x,\cdot) - \xi_{-}(\mathsf{X})K(x',\cdot) \Vert_{TV}\Vert f \Vert_{\infty}. \end{align*}