Jump Theorem for Cauchy Integrals:Suppose that $r$ is a smooth curve that passes through $z_0$, and $f(z)$ is a continuous function on $r$ that is analytic at $z_0$. Let $U$ be a small disk containing $z_0$ such that $f(z)$ is analytic on $U$ and such that $r$ divides $U$ into the two components $U_+$ and $U_-$ as above. $F(ζ)$ is cauchy integral of $f(z)$ over $r$. Then there are analytic functions $F_+(ζ)$ and $F_-(ζ)$ on $U$ satisfying :$F(ζ)$=$F_+(ζ)$ as $ζ∈U_+$,$F(ζ)$=$F_-(ζ)$ as $ζ∈U_-$, and $F_-(ζ)$=$F_+(ζ)+f(ζ)$as $ζ∈U$. $F_-(ζ)$=$\frac{1}{2πi} \int_{r+} \frac{f(z)}{z-ζ}dz +\frac{1}{2πi} \int_{r-r_0} \frac{f(z)}{z-ζ}dz$ $F_+(ζ)$=$\frac{1}{2πi} \int_{r-} \frac{f(z)}{z-ζ}dz +\frac{1}{2πi} \int_{r-r_0} \frac{f(z)}{z-ζ}dz$. $r-$ is left boundary of $U_-$,$r+$ is right boundary of $U_+$,$r_0$ is the common boundary of $U_+$ and $U_-$.
uniqueness of analytic function:if $f(z)$ and $g(z)$ is analytic function on $D$,and $f(z)=g(z)$ on a set which has a limitpoint,then $f(z)=g(z)$ on $D$.
Confuse:since $F(ζ)=F_+(ζ)$ as $ζ∈U_+$, $F(ζ)$ and $F_+(ζ)$ are analytic on ${U-r_0}$,so$F(ζ)=F_+(ζ)$ as $ζ∈\{U-r_0\}$,then $F_+(ζ)=F_-(ζ)$ as$ζ∈U_-$,this is contraduction with euqality $F_-(ζ)$=$F_+(ζ)+f(ζ)$as $ζ∈U$.
