Boundary vanishing of a nonzero Hardy function is confined to a null set

Part (a). The function $f(z)=1-z$ is a polynomial, hence holomorphic, with $\mathrm{\mid }f\mathrm{\mid }\le 2$ on $\mathbb{D}$, so $f\in H^{\mathrm{\infty }}(\mathbb{D})$; it is not identically zero. Its radial limits are $f^{\ast }(\zeta )=1-\zeta$, continuous on $\mathbb{T}$ and vanishing exactly at $\zeta =1$ by [F5], a set of measure zero.

Part (b). Let $f\in H^p$ with $\{f^{\ast }=0\}$ of positive measure. If $f\equiv ̸0$, then [F1] gives $\log \mathrm{\mid }{f}^{\ast }\mathrm{\mid }\in L^1$, so $f^{\ast }\mathrm{\ne }0$ $m$-almost everywhere, contradicting positive measure of the zero set; hence $f\equiv 0$.

Part (c). Let $f,g\in H^p$ with $f^{\ast }=g^{\ast }$ a.e. By [F2], the difference $f-g$ lies in $H^{p^{\mathrm{\prime }}}$ for some $0<p^{\mathrm{\prime }}\le \mathrm{\infty }$ (take $p^{\mathrm{\prime }}=p$); its boundary function $(f-g{)}^{\ast }=f^{\ast }-g^{\ast }$ vanishes a.e., a set of full measure. If $f-g\equiv ̸0$, then by step 1.2 applied to $f-g$ its zero set would have to be null, contradicting that it has full measure; hence $f=g$.

Part (d): $1-z$ is outer. By [F3] and [F4], $1-z=[\mathrm{\mid }1-\zeta \mathrm{\mid }]$; since $[h](0)=1>0$ and $e^{i\gamma }$ is normalized by the value at the origin, the unimodular constant is $1$. Hence $1-z$ is outer, and its canonical factorization $1-z=\lambda B{S}_{\mu }F$ has $B=1$ (no zeros in $\mathbb{D}$), $S_{\mu }=1$ (no singular factor for an outer function) and $F=1-z$ with $\lambda =1$.

Assembly. Step 1.1 proves (a), step 1.2 proves (b), step 2.1 proves the uniqueness statement (c), and step 2.2 identifies $1-z$ as the outer function $[\mathrm{\mid }1-\zeta \mathrm{\mid }]$ with the stated trivial canonical factors, proving (d).