The finite Holevo measurement bound

Fix $x$ with $p_x>0$. The operator $\bar{\rho }-p_x{\rho }_x={\sum }_{z\mathrm{\ne }x}{p}_z{\rho }_z$ is non-negative. If $v\in \ker \bar{\rho }$, then $0=⟨\bar{\rho }v,v⟩\ge p_x⟨{\rho }_{x}v,v⟩\ge 0$, so [F6] gives ${\rho }_{x}v=0$. Thus $\ker \bar{\rho }\subseteq \ker {\rho }_x$ and $\mathrm{supp}({\rho }_x)\subseteq \mathrm{supp}(\bar{\rho })$; $D({\rho }_x\mathrm{\parallel }\bar{\rho })$ is finite for every $x$ with $p_x>0$.

The measurement channel [F4] sends ${\rho }_x$ to the diagonal state with entries $w(y\mid x)$ and $\bar{\rho }$ to the diagonal state with entries $q_y$ by linearity [F7]. For $p_x>0$, if $w(y\mid x)>0$ then $q_y\ge p_{x}w(y\mid x)>0$. The QRE formula [F2] on diagonal states and data processing [F3] therefore give $D({\rho }_x\mathrm{\parallel }\bar{\rho })\ge {\sum }_{y:w(y\mid x)>0}w(y\mid x)\ln \frac{w(y\mid x)}{q_y}$.

Multiply step 2.1 by $p_x$ and sum only over $x$ with $p_x>0$. Extend the support logarithm of $\bar{\rho }$ by zero on its kernel; step 1.1 and [F2] ensure that this preserves every retained cross trace. Trace linearity [F7] and $\bar{\rho }={\sum }_{x:p_x>0}{p}_x{\rho }_x$ give ${\sum }_{x:p_x>0}{p}_{x}D({\rho }_x\mathrm{\parallel }\bar{\rho })={\sum }_{x:p_x>0}{p}_x\mathrm{tr}({\rho }_x\ln {\rho }_x)-\mathrm{tr}(\bar{\rho }\ln \bar{\rho })=\chi$ by [F1]. The summed measurement terms are exactly the displayed $I(X:Y)$ by its definition. Hence $I(X:Y)\le \chi$.