Solution
= Solution
Set $Z=g(X)$. Given $X$, the value of $Z$ is certain, so $H(Z\mid X)=0$. Applying the <Shannon entropy>[entropy] identity in the two orders gives
$$
H(X)=H(X,Z)=H(Z)+H(X\mid Z)\geq H(Z),
$$
because <conditional entropy> is nonnegative. Consequently
$$
\boxed{H(g(X))\leq H(X).}
$$
A deterministic coarsening can merge distinguishable source outcomes, but cannot create additional <Shannon entropy>.