Let $B$,$B_{1}$ and $B_{2}$ be non-interesecfting simple closed curves, such that $B_{1}$ and $B_{2}$ are contained inside $B$. Let $g$ be analytic in the region between these curves, and on the curves. I want to prove using only the Cauchy-Goursat theorem and basic properties of integrals that $$\int_Bg(z)dz=\int_{B_1}g(z)dz+\int_{B_2}g(z)dz$$
My attempt at the proof; First let the direction along $B$ be anti-clockwise, then since $g$ satisfies the conditions for the cauchy-goursat theorem we have that $$\int_{B}g(z)dz=0$$ Now let the direction along $B_{1}$ and $B_{2}$ be clockwise, then by the cauchy goursat theorem again we have that $$\int_{-B_{1}}g(z)dz+\int_{-B_{2}}g(z)dz=0\implies -\int_{B_1}g(z)dz-\int_{B_2}g(z)dz=0$$ $$\implies \int_{B_1}g(z)dz+\int_{B_2}g(z)dz=0=\int_{B}g(z)dz=\sum_{i=1}^{2}\int_{B_{i}}g(z)dz$$ as required
Could someone tell me if this is incorrect (and how to fix it if it is) and if it is correct?
This proposition as it is stated now does not hold. $B_1$ or $B_2$ is allowed to be inside of the other, resulting in $$\int_Bg(z)dz=\int_{B_1}g(z)dz=\int_{B_2}g(z)dz$$
Now we add the condition that neither of $B_1$ and $B_2$ resides inside of the other, as well as that $g$ is analytic in a (open and connected) domain $D$ that contains the union of the following two sets. The first set is the intersection of, the outerior of $B_1$ and $B_2$, and the interior of $B$. The second is $B\cup B_1\cup B_2$. This will produce the desired equation.
Proof: Transform the domain homeomorphically so that $B\cup B_1\cup B_2$ becomes the circles depicted in the graph below. Then take the contour $C$ as shown in the graph below running counter-clockwise on the large circle, clockwise on the smaller circles, only letting the distance between the gap be zero.
Take the contour $\gamma$ that is the image of $C$ on the original domain, apply Cauchy-Goursat theorem to $g$ on $D$ and contour $\gamma$ $$0=\int_\gamma g(z)dz=\int_B g(z)dz-\int_{B_1}g(z)dz-\int_{B_2}g(z)dz+0+0$$ where the last two $0$'s signifies the integration on the parallel sides of the straight "streets" in the graph in opposite directions which results in cancellation of the integration on those parts of $\gamma$.
Now we have obtained the desired equation.