[proofplan] We define $F$ by reflecting: $F(z) = \overline{f(\bar{z})}$ for $z \in \Omega^-$. We first show $F$ is holomorphic on $\Omega^-$ by verifying the Cauchy--Riemann equations for the conjugate-reflect construction. Continuity on $I$ follows from the hypothesis that $f$ is real-valued there, which forces the two definitions to agree. Finally, we apply Morera's theorem to prove holomorphicity on the full domain $\Omega^+ \cup I \cup \Omega^-$ by showing that contour integrals over triangles crossing $I$ vanish. [/proofplan]

[step:Define $F$ and verify it is holomorphic on $\Omega^-$]Define $F: \Omega^+ \cup I \cup \Omega^- \to \mathbb{C}$ by \begin{align*} F(z) &= \begin{cases} f(z) & z \in \Omega^+ \cup I \\ \overline{f(\bar{z})} & z \in \Omega^-. \end{cases} \end{align*} We show $F$ is holomorphic on $\Omega^-$. Define the auxiliary function $g: \Omega^- \to \mathbb{C}$ by $g(z) = \overline{f(\bar{z})}$. The map $z \mapsto \bar{z}$ sends $\Omega^-$ biholomorphically onto $\Omega^+$ (it is an antiholomorphic involution, and composing with conjugation restores holomorphicity). Explicitly: if $f = u + iv$ where $u, v: \Omega^+ \to \mathbb{R}$ are the real and imaginary parts, then for $z = x + iy \in \Omega^-$ (so $y < 0$ and $\bar{z} = x - iy \in \Omega^+$), \begin{align*} g(z) = \overline{f(\bar{z})} = u(x, -y) - iv(x, -y). \end{align*} Define $U(x, y) = u(x, -y)$ and $V(x, y) = -v(x, -y)$, so $g = U + iV$. Since $f$ is holomorphic on $\Omega^+$, the functions $u$ and $v$ satisfy the Cauchy--Riemann equations there: $\partial_x u = \partial_y v$ and $\partial_y u = -\partial_x v$. We verify the Cauchy--Riemann equations for $U$ and $V$ at each point $(x, y)$ with $y < 0$: \begin{align*} \partial_x U(x,y) &= \partial_x u(x, -y) = \partial_y v(x, -y) = -\partial_y(-v(x,-y)) = \partial_y V(x,y), \\ \partial_y U(x,y) &= -\partial_y u(x,-y)\big|_{\text{chain rule}} = -(-\partial_x v(x,-y)) = \partial_x v(x,-y) = -\partial_x V(x,y). \end{align*} Since $u$ and $v$ are $C^1$ on $\Omega^+$ (as real and imaginary parts of a holomorphic function), $U$ and $V$ are $C^1$ on $\Omega^-$. By the Cauchy--Riemann equations with $C^1$ regularity, $g$ is holomorphic on $\Omega^-$.[/step]

[guided]The idea is that conjugation is antiholomorphic: if $h$ is holomorphic, then $\bar{h}$ is antiholomorphic. But the reflection $z \mapsto \bar{z}$ is also antiholomorphic. Composing two antiholomorphic operations gives a holomorphic one. So $z \mapsto \overline{f(\bar{z})}$ is holomorphic on $\Omega^-$. To verify this rigorously, we write $f = u + iv$ and compute in real coordinates. For $z = x + iy \in \Omega^-$, we have $y < 0$, so $\bar{z} = x + i(-y) \in \Omega^+$. Then: \begin{align*} g(z) = \overline{f(\bar{z})} = \overline{u(x,-y) + iv(x,-y)} = u(x,-y) - iv(x,-y). \end{align*} Setting $U(x,y) := u(x,-y)$ and $V(x,y) := -v(x,-y)$, we need to check $\partial_x U = \partial_y V$ and $\partial_y U = -\partial_x V$. For the first equation, the chain rule gives $\partial_x U(x,y) = \partial_x u(x,-y)$. Since $u, v$ satisfy Cauchy--Riemann on $\Omega^+$, we have $\partial_x u(x,-y) = \partial_y v(x,-y)$. Meanwhile $\partial_y V(x,y) = -\partial_y v(x,-y) \cdot (-1) = \partial_y v(x,-y)$ by the chain rule (differentiating $v(x, -y)$ with respect to $y$ introduces a factor of $-1$). So $\partial_x U = \partial_y V$. For the second equation, $\partial_y U(x,y) = \partial_y u(x,-y) \cdot (-1) = -\partial_y u(x,-y)$. By Cauchy--Riemann, $\partial_y u(x,-y) = -\partial_x v(x,-y)$, so $\partial_y U = \partial_x v(x,-y)$. Meanwhile $-\partial_x V(x,y) = -(-\partial_x v(x,-y)) = \partial_x v(x,-y)$. So $\partial_y U = -\partial_x V$. Since $f$ is holomorphic, $u$ and $v$ are real-analytic (in particular $C^1$), so $U$ and $V$ are $C^1$ on $\Omega^-$. The Cauchy--Riemann equations with continuous partials imply $g$ is holomorphic on $\Omega^-$.[/guided]

[step:Verify continuity and agreement on $I$]For $x_0 \in I$, we must show the two definitions of $F$ agree and that $F$ is continuous at $x_0$. From above on $\Omega^+ \cup I$, $F(x_0) = f(x_0)$. For the formula on $\Omega^-$, the limit as $z \to x_0$ from $\Omega^-$ gives $\lim_{z \to x_0, z \in \Omega^-} \overline{f(\bar{z})}$. Since $z \to x_0$ with $z \in \Omega^-$ implies $\bar{z} \to x_0$ with $\bar{z} \in \Omega^+$, and $f$ is continuous on $\Omega^+ \cup I$, we get \begin{align*} \lim_{z \to x_0,\, z \in \Omega^-} \overline{f(\bar{z})} = \overline{f(x_0)}. \end{align*} Since $f(x_0) \in \mathbb{R}$ (by hypothesis that $f$ is real-valued on $I$), we have $\overline{f(x_0)} = f(x_0) = F(x_0)$. Therefore $F$ is continuous on $\Omega^+ \cup I \cup \Omega^-$.[/step]

[guided]Why do the two formulas match on $I$? The key hypothesis is that $f$ is real-valued on $I$. For $x_0 \in I$, the "upper" definition gives $F(x_0) = f(x_0) \in \mathbb{R}$. The "lower" formula, evaluated by continuity at $x_0$, gives $\overline{f(\overline{x_0})} = \overline{f(x_0)}$. Since $f(x_0)$ is real, $\overline{f(x_0)} = f(x_0)$. Without the real-valuedness condition, the two definitions would generally disagree on $I$, and $F$ would have a jump discontinuity there. More precisely: $F$ is continuous on $\Omega^+$ (where it equals $f$, holomorphic hence continuous), on $\Omega^-$ (where it equals $g$, holomorphic hence continuous), and the limits from both sides agree at every point of $I$ with the value $f(x_0) = F(x_0)$. So $F$ is continuous on the entire domain $\Omega^+ \cup I \cup \Omega^-$.[/guided]

[step:Apply Morera's theorem to establish holomorphicity across $I$]We have established that $F$ is holomorphic on $\Omega^+$ and on $\Omega^-$, and continuous on the full domain $\Omega := \Omega^+ \cup I \cup \Omega^-$. It remains to show $F$ is holomorphic on $\Omega$; the only points requiring proof are those on $I$. We apply [Morera's Theorem](/theorems/???): it suffices to show that $\oint_{\partial T} F(z)\, dz = 0$ for every triangle $T$ with $\overline{T} \subset \Omega$. If $T \subset \Omega^+$ or $T \subset \Omega^-$, then $F$ is holomorphic on a neighbourhood of $\overline{T}$, and [Cauchy's Theorem](/theorems/???) gives $\oint_{\partial T} F(z)\, dz = 0$. Now suppose $T$ intersects $I$. For $\varepsilon > 0$, define the shifted segment $I_\varepsilon = \{z \in T : \operatorname{Im}(z) = \varepsilon\}$ and $I_{-\varepsilon} = \{z \in T : \operatorname{Im}(z) = -\varepsilon\}$. Split the boundary integral into three parts: \begin{align*} \oint_{\partial T} F(z)\, dz &= \int_{\partial T \cap \overline{\{\operatorname{Im}(z) \geq \varepsilon\}}} F(z)\, dz + \int_{\partial T \cap \overline{\{|\operatorname{Im}(z)| \leq \varepsilon\}}} F(z)\, dz + \int_{\partial T \cap \overline{\{\operatorname{Im}(z) \leq -\varepsilon\}}} F(z)\, dz. \end{align*} More precisely, define the upper sub-triangle $T^+_\varepsilon := T \cap \{\operatorname{Im}(z) > \varepsilon\}$ and lower sub-region $T^-_\varepsilon := T \cap \{\operatorname{Im}(z) < -\varepsilon\}$. By Cauchy's theorem applied to $F$ (holomorphic on $\Omega^+$ and $\Omega^-$ respectively), the contour integrals over $\partial T^+_\varepsilon$ and $\partial T^-_\varepsilon$ vanish. The remaining contribution comes from the strip $\{z \in T : |\operatorname{Im}(z)| \leq \varepsilon\}$, whose contour integral is bounded by \begin{align*} \left| \int_{\text{strip boundary}} F(z)\, dz \right| &\leq \sup_{z \in \overline{T}} |F(z)| \cdot \operatorname{length}(\text{strip boundary} \cap \partial T) \cdot C\varepsilon \end{align*} for a geometric constant $C$ depending on $T$. Since $F$ is continuous on the compact set $\overline{T}$, it is bounded there. As $\varepsilon \to 0$, the strip contribution vanishes, giving \begin{align*} \oint_{\partial T} F(z)\, dz = 0. \end{align*} By Morera's theorem, $F$ is holomorphic on $\Omega^+ \cup I \cup \Omega^-$.[/step]

[guided]The difficulty is proving holomorphicity on $I$ itself. We know $F$ is holomorphic away from $I$ and continuous across $I$. Morera's theorem converts the holomorphicity question into a vanishing-integral question: if $F$ is continuous on an open set $\Omega$ and $\oint_{\partial T} F\, dz = 0$ for every triangle $T \subset \Omega$, then $F$ is holomorphic on $\Omega$. For triangles entirely contained in $\Omega^+$ or $\Omega^-$, Cauchy's theorem applies directly since $F$ is holomorphic there. The interesting case is a triangle $T$ that straddles $I$. The strategy is to cut $T$ along a horizontal line at height $\varepsilon$ (and $-\varepsilon$), apply Cauchy's theorem to each piece that lies entirely in $\Omega^+$ or $\Omega^-$, and show the leftover integral (along paths near $I$) tends to zero as $\varepsilon \to 0$ by the continuity and boundedness of $F$ on $\overline{T}$. Concretely: the triangle $T$ is split into at most three regions by the lines $\operatorname{Im}(z) = \pm\varepsilon$. The upper and lower regions have boundary integrals that vanish by Cauchy's theorem. The middle strip has perimeter $O(\varepsilon)$ (its height is $2\varepsilon$ and its horizontal extent is bounded by the diameter of $T$), so the integral over its boundary is at most $\|F\|_\infty \cdot O(\varepsilon) \to 0$. Summing and taking $\varepsilon \to 0$ gives $\oint_{\partial T} F\, dz = 0$. Since $\Omega^+ \cup I \cup \Omega^-$ is open (as $I$ is an open segment of $\mathbb{R}$, and $\Omega^+$, $\Omega^-$ are open), Morera's theorem applies and yields holomorphicity of $F$ on the entire domain.[/guided]