[proofplan] Starting from the defining sum $\operatorname{Ind}^G_H \psi(g) = \frac{1}{|H|}\sum_{x \in G}\mathring{\psi}(x^{-1}gx)$, only the terms with $x^{-1}gx \in H$ contribute. The set $S_g = \{x \in G : x^{-1}gx \in H\}$ is empty exactly when no $G$-conjugate of $g$ lies in $H$, i.e., when $\mathcal{C}_G(g) \cap H = \varnothing$ — the case $m = 0$. When $m \geq 1$, we partition $S_g$ according to which $H$-conjugacy class $x^{-1}gx$ falls into. Each piece $X_i = \{x : x^{-1}gx \sim_H x_i\}$ is a single double coset $C_G(g)\, g_i\, H$ (where $g_i^{-1} g g_i = x_i$ realises the class membership), and the product formula $|C_G(g)\, g_i\, H| = |C_G(g)|\cdot|H|/|C_H(x_i)|$ — based on $g_i^{-1} C_G(g) g_i \cap H = C_H(x_i)$ — gives the size of each piece. Summing the contributions and dividing by $|H|$ produces the formula. [/proofplan]

[step:Reduce the defining sum to terms with $x^{-1} g x \in H$ and dispatch the case $m = 0$] By the defining formula for $\operatorname{Ind}^G_H \psi(g)$, \begin{align*} \operatorname{Ind}^G_H \psi(g) = \frac{1}{|H|} \sum_{x \in G} \mathring{\psi}(x^{-1} g x). \end{align*} The zero extension $\mathring{\psi}: G \to \mathbb{C}$ vanishes outside $H$, so terms with $x^{-1} g x \in G \setminus H$ contribute zero. Define \begin{align*} S_g := \{x \in G : x^{-1} g x \in H\}. \end{align*} Then \begin{align*} \operatorname{Ind}^G_H \psi(g) = \frac{1}{|H|} \sum_{x \in S_g} \psi(x^{-1} g x). \end{align*} Note $x \in S_g$ if and only if some $H$-element is a $G$-conjugate of $g$, i.e., $\mathcal{C}_G(g) \cap H \neq \varnothing$ (where $\mathcal{C}_G(g) = \{y g y^{-1} : y \in G\}$ is the $G$-conjugacy class of $g$). Equivalently, $S_g \neq \varnothing$ iff $\mathcal{C}_G(g) \cap H \neq \varnothing$. **Case $m = 0$.** If $\mathcal{C}_G(g) \cap H = \varnothing$, then $S_g = \varnothing$ and the sum is empty, giving $\operatorname{Ind}^G_H \psi(g) = 0$ as claimed. In what follows, assume $m \geq 1$. [/step]

[step:Partition $S_g$ by $H$-conjugacy class and identify the $\psi$-contribution of each part] By hypothesis, \begin{align*} \mathcal{C}_G(g) \cap H = \bigsqcup_{i=1}^m \mathcal{C}_H(x_i), \end{align*} where $x_1, \ldots, x_m$ are representatives of the $H$-conjugacy classes inside $\mathcal{C}_G(g) \cap H$ (the union is disjoint by definition of conjugacy classes as equivalence classes). Define \begin{align*} X_i := \{x \in G : x^{-1} g x \in \mathcal{C}_H(x_i)\}, \qquad i = 1, \ldots, m. \end{align*} Since each $x \in S_g$ has $x^{-1} g x \in \mathcal{C}_G(g) \cap H$, and the $\mathcal{C}_H(x_i)$ partition this intersection, we have \begin{align*} S_g = \bigsqcup_{i=1}^m X_i. \end{align*} For $x \in X_i$, $x^{-1} g x$ is $H$-conjugate to $x_i$, and $\psi$ is a class function of $H$, so \begin{align*} \psi(x^{-1} g x) = \psi(x_i). \end{align*} Hence \begin{align*} \operatorname{Ind}^G_H \psi(g) = \frac{1}{|H|} \sum_{i=1}^m \sum_{x \in X_i} \psi(x_i) = \frac{1}{|H|} \sum_{i=1}^m |X_i|\, \psi(x_i). \end{align*} The remaining task is to compute $|X_i|$. [/step]

[step:Identify $X_i$ as the double coset $C_G(g)\, g_i\, H$] Fix $i \in \{1, \ldots, m\}$. Since $x_i \in \mathcal{C}_G(g) \cap H \subseteq \mathcal{C}_G(g)$, there exists $g_i \in G$ with \begin{align*} g_i^{-1} g\, g_i = x_i. \end{align*} Such a $g_i$ exists by definition of the $G$-conjugacy class $\mathcal{C}_G(g)$, but is not unique — we fix one choice. **Inclusion $C_G(g)\, g_i\, H \subseteq X_i$.** Let $c \in C_G(g)$ and $h \in H$. We compute \begin{align*} (c g_i h)^{-1}\, g\, (c g_i h) = h^{-1} g_i^{-1} c^{-1}\, g\, c\, g_i h = h^{-1} g_i^{-1} g\, g_i h = h^{-1} x_i h, \end{align*} using $c^{-1} g c = g$ (since $c$ commutes with $g$) and $g_i^{-1} g\, g_i = x_i$. Now $h^{-1} x_i h \in \mathcal{C}_H(x_i)$, so $c g_i h \in X_i$. **Inclusion $X_i \subseteq C_G(g)\, g_i\, H$.** Let $x \in X_i$. By definition, $x^{-1} g x \in \mathcal{C}_H(x_i)$, so there exists $h \in H$ with \begin{align*} x^{-1} g x = h^{-1} x_i h = h^{-1} g_i^{-1} g\, g_i h. \end{align*} Rearranging, \begin{align*} g\, x = x\, h^{-1} g_i^{-1} g\, g_i h \qquad \implies \qquad g\, (x h^{-1} g_i^{-1}) = (x h^{-1} g_i^{-1})\, g. \end{align*} That is, $x h^{-1} g_i^{-1}$ commutes with $g$, hence lies in $C_G(g)$. Set $c := x h^{-1} g_i^{-1} \in C_G(g)$. Then \begin{align*} x = c\, g_i\, h \in C_G(g)\, g_i\, H. \end{align*} Combining the two inclusions, $X_i = C_G(g)\, g_i\, H$. [/step]

[step:Compute $|X_i|$ via the double-coset product formula]The size of a double coset $A g_i B$ for subgroups $A, B \leq G$ is given by the product formula \begin{align*} |A\, g_i\, B| = \frac{|A|\, |B|}{|A \cap g_i B g_i^{-1}|} \end{align*} (equivalently $= |A|\,|B|/|g_i^{-1} A g_i \cap B|$, which is the form we use here). With $A = C_G(g)$ and $B = H$, \begin{align*} |C_G(g)\, g_i\, H| = \frac{|C_G(g)|\, |H|}{|g_i^{-1} C_G(g) g_i \cap H|}. \end{align*} We now identify $g_i^{-1} C_G(g) g_i \cap H = C_H(x_i)$. **Identification $g_i^{-1} C_G(g) g_i = C_G(x_i)$.** For $c \in G$, $c \in C_G(g)$ iff $cg = gc$ iff $g_i^{-1} c g_i \cdot g_i^{-1} g g_i = g_i^{-1} g g_i \cdot g_i^{-1} c g_i$, iff (using $g_i^{-1} g g_i = x_i$) the conjugate $g_i^{-1} c g_i$ commutes with $x_i$, iff $g_i^{-1} c g_i \in C_G(x_i)$. So the conjugation map $c \mapsto g_i^{-1} c g_i$ takes $C_G(g)$ bijectively onto $C_G(x_i)$: \begin{align*} g_i^{-1} C_G(g) g_i = C_G(x_i). \end{align*} **Intersection $C_G(x_i) \cap H = C_H(x_i)$.** For $z \in G$, $z \in C_G(x_i) \cap H$ iff $z \in H$ and $z$ commutes with $x_i$ — exactly the definition of $C_H(x_i)$. So \begin{align*} g_i^{-1} C_G(g) g_i \cap H = C_G(x_i) \cap H = C_H(x_i). \end{align*} Substituting, \begin{align*} |X_i| = |C_G(g)\, g_i\, H| = \frac{|C_G(g)|\, |H|}{|C_H(x_i)|}. \end{align*}[/step]

[guided]The double-coset product formula is the algebraic engine here. To remember it: $A g_i B$ is a union of left cosets of $B$, indexed by elements of $A / (A \cap g_i B g_i^{-1})$. Each left coset has $|B|$ elements, and the index is $|A|/|A \cap g_i B g_i^{-1}|$. Multiplying gives $|A\, g_i\, B| = |A|\,|B|/|A \cap g_i B g_i^{-1}|$. In the form we need, we conjugate the denominator: $|A \cap g_i B g_i^{-1}| = |g_i^{-1}(A \cap g_i B g_i^{-1})g_i| = |g_i^{-1} A g_i \cap B|$, since conjugation by $g_i^{-1}$ is a bijection. **Why $g_i^{-1} C_G(g) g_i = C_G(x_i)$.** Because conjugation by $g_i$ takes $g$ to $g_i g g_i^{-1}$ — wait, we set $g_i^{-1} g g_i = x_i$, so $g = g_i x_i g_i^{-1}$, i.e., conjugation by $g_i$ takes $x_i$ to $g$. Centralisers transform covariantly under conjugation: if $\sigma$ is the conjugation map $c \mapsto g_i^{-1} c g_i$, then $\sigma(C_G(g)) = C_G(\sigma^{-1}(g)) = C_G(g_i g g_i^{-1})$... actually, the cleanest statement is that for any $w \in G$, $C_G(w g w^{-1}) = w\, C_G(g)\, w^{-1}$. With $w = g_i^{-1}$ (since $g_i^{-1} g g_i^{-(-1)} = g_i^{-1} g g_i$ — check the bracketing): we want $C_G(g_i^{-1} g g_i)$, which equals $g_i^{-1} C_G(g) g_i$. So yes, $g_i^{-1} C_G(g) g_i = C_G(x_i)$. **Why pick a single $g_i$ and not worry about the choice?** The double coset $C_G(g)\, g_i\, H$ depends only on the $H$-conjugacy class of $g_i^{-1} g g_i = x_i$, not on the specific choice of $g_i$. If $g_i'$ is another conjugating element with $(g_i')^{-1} g\, g_i' = x_i$, then $g_i (g_i')^{-1}$ centralises $g$ (so it is in $C_G(g)$), and $g_i' = g_i \cdot (g_i^{-1} g_i')$ where $g_i^{-1} g_i' \in C_G(g)\, H$ — i.e., $g_i'$ lies in the same double coset as $g_i$, so the double coset $C_G(g) g_i H = C_G(g) g_i' H$ is unchanged.[/guided]

[step:Assemble the formula] Combining Steps 2 and 4, \begin{align*} \operatorname{Ind}^G_H \psi(g) = \frac{1}{|H|} \sum_{i=1}^m |X_i|\, \psi(x_i) = \frac{1}{|H|} \sum_{i=1}^m \frac{|C_G(g)|\,|H|}{|C_H(x_i)|}\, \psi(x_i). \end{align*} The factor $|H|$ cancels, leaving \begin{align*} \operatorname{Ind}^G_H \psi(g) = |C_G(g)| \sum_{i=1}^m \frac{\psi(x_i)}{|C_H(x_i)|}, \end{align*} as claimed. Together with the case $m = 0$ from Step 1, this completes the proof. [/step]