Let $P_\pm$ be the Cauchy projectors defined by their action on $C^1(S^1)$ function $f$, where $S^1 = \{ \zeta \in \mathbb C \mid |z| = 1 \}$, by the rule $$ P_\pm f(z) = \lim\limits_{\varepsilon \to + 0} \frac{1}{2\pi i} \int\limits_{|\zeta|=1}\frac{f(\zeta)}{\zeta - z\cdot(1\mp\varepsilon)}d\zeta, \quad |z| = 1. $$ So $P_\pm$ send functions on $S^1$ to functions on $S^1$. They satisfy two equalities: \begin{align} \int\limits_{|z|=1}P_+f(z) \, dz &=0, \tag{1}\\ \int\limits_{|z|=1}P_- f(z)\, dz &= -\int\limits_{|z|=1}f(z) \, dz.\tag{2} \end{align} My question is how to prove these equalities.
The function $$ \varphi_+(z) = \frac{1}{2\pi i}\int\limits_{|z|=1}\frac{f(\zeta)}{\zeta - z} \, d\zeta, \quad |z| < 1, $$ is holomorphic in the open ball $\{ z \in \mathbb C \mid |z| < 1 \}$. Then for any small $\varepsilon > 0$ the function $z \mapsto \varphi_+(z\cdot(1-\varepsilon))$ is holomorphic in the open ball $\{ z \in \mathbb C \mid |z| < (1-\varepsilon)^{-1} \}$. Hence for any small $\varepsilon > 0$ we have by the Cauchy formula $$ \int\limits_{|z|=1} \varphi_+(z\cdot(1-\varepsilon)) \, dz = 0 $$ and $$ \lim\limits_{\varepsilon \to + 0}\int\limits_{|z|=1} \varphi_+(z\cdot(1-\varepsilon)) \, dz = 0. $$ Now take into account that $\lim_{\varepsilon \to + 0} \varphi_+(z \cdot (1-\varepsilon)) = P_+ f(z)$ for $z \in S^1$. So to prove the formula (1) it is sufficient to show that $$ \lim\limits_{\varepsilon \to + 0}\int\limits_{|z|=1} \varphi_+(z\cdot(1-\varepsilon)) \, dz = \int\limits_{|z|=1} \lim\limits_{\varepsilon \to + 0} \varphi_+(z\cdot(1-\varepsilon)) \, dz. \tag{3} $$ If this approach is correct then the formula (2) can be proved using the same method. So the question is why the formula (3) is valid.