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Theorem 6.5 from Garnett's 2007 book Bounded Analytic Functions is as follows. I quote it verbatim, because I am concerned about the possibility of there being a typo:

Theorem 6.5. Let $v\left(z\right)$ be a subharmonic function in the unit disk $D$. Assume $v(z)\not\equiv-\infty$. For $0<r<1$, let $$v_{r}(z)=\begin{cases} v(z), & \lvert z\rvert\leq r,\\ \frac{1}{2\pi}\int P_{z/r}(\theta)v(re^{i\theta})\mathrm d\theta, & \lvert z\rvert<r. \end{cases}$$ Then $v_{r}(z)$ is a subharmonic function in $D$, $v_{r}(z)$ is harmonic on $\lvert z\rvert<r$, $v_{r}(z)\geq v(z)$, $z\in D$, and $v_{r}(z)$ is an increasing function of $r$.

Am I correct in assuming that the second condition $\lvert z\rvert<r$ in the definition of $v_{r}(z)$ is incorrect, and that it should read $r<\lvert z\rvert<1$? If so, please include "Yes, you are correct" in your response. If am not correct, please include "No, you are incorrect" in your response and tell me exactly how I am supposed to interpret this result. This detail is very important for my research, so I want to make sure I get it exactly right.

LSpice
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MCS
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    I bet that the first condition ("$|z| \leqslant r$") should read "$r \leqslant |z| < 1$". – Mateusz Kwaśnicki Dec 15 '20 at 22:03
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    you are not correct as the second part is correct since that implies by the subharmonicity of $v$ that $v(z) \le v_r(z), |z|<r$ (which is what we want from later text) hence the first part must be indeed $v_r=v, r \le |z| <1$ as @Mateusz also noted – Conrad Dec 15 '20 at 22:17
  • Thanks a bunch! – MCS Dec 15 '20 at 22:20

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