[proofplan] We express $S_N f(x) - s$ as a convolution integral against the Dirichlet kernel, subtract $s$ using the unit-mass property $\frac{1}{2\pi}\int D_N = 1$, symmetrise the integral to $(0,\pi)$, and factor the kernel as $\sin((N+\tfrac{1}{2})t)/\sin(t/2)$. Dividing by $2\sin(t/2)$ produces a quotient function $g$ whose integrability is guaranteed by the Dini condition. The [Riemann-Lebesgue Lemma](/theorems/245) applied to $g$ forces $\int g(t)\sin((N+\tfrac{1}{2})t) \to 0$. [/proofplan] [step:Write $S_N f(x) - s$ as a symmetrised integral against $D_N$] By the [Dirichlet Kernel Formula](/theorems/581), $S_N f(x) = \frac{1}{2\pi}\int_{-\pi}^\pi f(x-t)\,D_N(t) \, d\mathcal{L}^1(t)$. Since $\frac{1}{2\pi}\int_{-\pi}^\pi D_N(t) \, d\mathcal{L}^1(t) = 1$ (the $n = 0$ term contributes $1$ and all other terms integrate to $0$): \begin{align*} S_N f(x) - s = \frac{1}{2\pi}\int_{-\pi}^\pi [f(x-t) - s]\,D_N(t) \, d\mathcal{L}^1(t). \end{align*} Split over $[-\pi, 0]$ and $[0, \pi]$. In the integral over $[-\pi, 0]$, substitute $t \mapsto -t$ and use the evenness of $D_N$ (from the closed form $\sin((N+\tfrac{1}{2})t)/\sin(t/2)$, which is a ratio of odd functions and hence even): \begin{align*} S_N f(x) - s = \frac{1}{2\pi}\int_0^\pi [f(x+t) + f(x-t) - 2s]\,D_N(t) \, d\mathcal{L}^1(t). \end{align*} [/step] [step:Factor the Dirichlet kernel and define the quotient function $g$] By the [Dirichlet Kernel Formula](/theorems/581), $D_N(t) = \sin((N+\tfrac{1}{2})t)/\sin(t/2)$ for $t \notin 2\pi\mathbb{Z}$. Substituting: \begin{align*} S_N f(x) - s = \frac{1}{2\pi}\int_0^\pi \frac{f(x+t) + f(x-t) - 2s}{\sin(t/2)} \cdot \sin\!\left(\left(N + \tfrac{1}{2}\right)t\right) d\mathcal{L}^1(t). \end{align*} Define the quotient function: \begin{align*} g: (0, \pi] &\to \mathbb{R} \\ t &\mapsto \frac{f(x+t) + f(x-t) - 2s}{2\sin(t/2)}. \end{align*} Then: \begin{align*} S_N f(x) - s = \frac{1}{\pi}\int_0^\pi g(t)\,\sin\!\left(\left(N + \tfrac{1}{2}\right)t\right) d\mathcal{L}^1(t). \end{align*} [claim:The Dini condition implies $g \in L^1((0,\pi))$] [/claim] [proof] The concavity bound $\sin\theta \geq 2\theta/\pi$ for $\theta \in [0, \pi/2]$ gives $\sin(t/2) \geq t/\pi$ for $t \in [0, \pi]$. Therefore: \begin{align*} |g(t)| = \frac{|f(x+t) + f(x-t) - 2s|}{2\sin(t/2)} \leq \frac{\pi}{2} \cdot \frac{|f(x+t) + f(x-t) - 2s|}{t}. \end{align*} Integrating: \begin{align*} \int_0^\pi |g(t)| \, d\mathcal{L}^1(t) \leq \frac{\pi}{2}\int_0^\pi \frac{|f(x+t) + f(x-t) - 2s|}{t} \, d\mathcal{L}^1(t) < \infty \end{align*} by the Dini condition. [/proof] [/step] [step:Apply the Riemann--Lebesgue lemma to conclude $S_N f(x) \to s$] For any $h \in L^1((0,\pi))$, the integral $\int_0^\pi h(t)\,\sin(\omega t) \, d\mathcal{L}^1(t) \to 0$ as $\omega \to \infty$. This follows from the [Riemann-Lebesgue Lemma](/theorems/245): extend $h$ by zero to $\mathbb{R}$, write $\int h\,\sin(\omega t) = \operatorname{Im}(\hat{h}(-\omega))$, and use $\hat{h}(\xi) \to 0$ as $|\xi| \to \infty$. Applying this with $h = g \in L^1((0,\pi))$ and $\omega = N + \tfrac{1}{2} \to \infty$: \begin{align*} S_N f(x) - s = \frac{1}{\pi}\int_0^\pi g(t)\,\sin\!\left(\left(N + \tfrac{1}{2}\right)t\right) d\mathcal{L}^1(t) \to 0. \end{align*} Therefore $S_N f(x) \to s$. [/step]