I had some difficulties understanding why a certain step in a proof works. Specifically in the proof of Schwarz's reflection principle using Morera's theorem.
Let $f$ be holomorphic on the open upper half plane, continuous on the closed upper half plane. Then we seek to prove that $\int_{\partial T}f(z)dz$ is zero, where $T\subset\overline{\mathbb{H}}$ is any closed triangle. It is clear in the case where the triangle completely lies on the upper half plane. The problem is when an edge of $T$ intersects the real axis, it is claimed that if we take $T_\epsilon$ to be the triangle obtained by raising the said edge of $T$ by $\epsilon$, then $\int_{\partial T_\epsilon} f(z)dz$ is clearly $0$. The result follows if we take $\epsilon\to 0$.
The part that confused me is why exactly is it true that $$\lim_{\epsilon\to 0}\int_{\partial T_\epsilon}f(z)dz=\int_{\lim_{\epsilon\to 0}\partial T_\epsilon}f(z)dz.$$ It seems very intuitive but then I couldn't come up with an epsilon delta proof that shows that it is true. An extension to this question is, under what assumption of $f$ would the above equality of limit be true.
Hint: Let $\gamma : [a,b] \to \overline{\mathbb H}$ be a contour. For $(\epsilon,t)\in [0,1]\times [a,b],$ define
$$\Gamma(\epsilon,t) = i\epsilon+\gamma(t).$$
Use the uniform continuity of $f$ on $\Gamma([0,1]\times [a,b])$ to see
$$\int_\gamma f(z)\,dz = \lim_{\epsilon\to 0^+}\int_{i\epsilon+\gamma} f(z)\,dz.$$