I am looking at the proof of the Symmetry Principle from Stein and Shakarchi's Complex Analysis. Here, we try to prove the theorem using Morera's theorem. My question here is regarding the argument given below in diagram (a).
So, from Morera's theorem, we have that $\int_{T_\epsilon} f=0$. We then let $\epsilon \to 0$, and conclude that $\int_T f(z)dz =0$. It says that we have this convergence by continuity. However, while this is intuitively clear, I can't think of a rigorous argument that guarantees $$\int_{T_\epsilon} f(z)dz = \int_T f(z)dz.$$ How do we get this result by continuity?
$T$ and $T_\epsilon$ are two piecewise $C^1$ curves
$$\gamma,\gamma_\epsilon:[a,b]\longrightarrow\Bbb C$$ with $\gamma_\epsilon = \gamma + i\epsilon.$ You must prove that the integrand of $$ \int_{T_\epsilon}f(z)dz = \int_a^b f(\gamma_\epsilon(t))\gamma_\epsilon'(t)dt $$ converges uniformly to the integrand of $$ \int_{T}f(z)dz = \int_a^b f(\gamma(t))\gamma'(t)dt $$ (the derivatives are undefined in two points)