= Solution
{c}
Using the <Cauchy distribution> from the preceding part gives one common <probability space> for all starting points:
$$
\phi(x+iy)=\mathbb E[f(x+yC)]\quad(y>0).
$$
As $(x,y)\to(0,0)$ with $y>0$, $x+yC\to0$ almost surely, since $C$ is a finite real <random variable> almost surely. <Continuity> of $f$ at zero gives $f(x+yC)\to f(0)$ almost surely, and boundedness supplies an integrable constant bound. <Dominated convergence theorem> therefore proves
$$
\boxed{\lim_{\substack{z\to0\\\Im z>0}}\phi(z)=f(0)=\phi(0).}
$$
The argument applies to every sequence approaching zero, including tangential approaches. Along the real boundary the extension is already $f$, which is continuous. Hence the extension is \b[continuous at zero in the closed half-plane]. No finite moment of the <Cauchy distribution> is required.
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