[proofplan]
We substitute the definition of $\tilde{f}(k)$ into the inversion integral, interchange the order of integration (justified by absolute integrability of $f$), and recognise the inner integral as a Dirichlet kernel. The Dirichlet kernel representation reduces the problem to showing that the symmetric partial integrals converge to $\frac{1}{2}(f(x^+) + f(x^-))$, which follows from the Riemann-Lebesgue lemma applied to the localised integral.
[/proofplan]
[step:Substitute the definition of $\tilde{f}$ and interchange the order of integration]
Define the symmetric partial integral $S_R(x) = \frac{1}{2\pi}\int_{-R}^{R} \tilde{f}(k)e^{ikx}\,d\mathcal{L}^1(k)$. Substituting the definition $\tilde{f}(k) = \int_{-\infty}^{\infty} f(y)e^{-iky}\,d\mathcal{L}^1(y)$:
\begin{align*}
S_R(x) = \frac{1}{2\pi}\int_{-R}^{R}\left(\int_{-\infty}^{\infty} f(y)e^{-iky}\,d\mathcal{L}^1(y)\right)e^{ikx}\,d\mathcal{L}^1(k).
\end{align*}
Since $f \in L^1(\mathbb{R})$ and the $k$-integral is over the finite interval $[-R, R]$, the iterated integral is absolutely convergent. By Fubini's theorem, we interchange the order:
\begin{align*}
S_R(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} f(y)\left(\int_{-R}^{R} e^{ik(x - y)}\,d\mathcal{L}^1(k)\right)d\mathcal{L}^1(y).
\end{align*}
[guided]
We want to show $\lim_{R \to \infty} S_R(x) = \frac{1}{2}(f(x^+) + f(x^-))$. The strategy is to express $S_R(x)$ as $f$ convolved with a kernel (the Dirichlet kernel), then analyse the kernel's asymptotic behaviour as $R \to \infty$.
Substituting the definition of $\tilde{f}$ gives a double integral in $(y, k)$. To interchange the order of integration, we verify absolute integrability of the integrand over $\mathbb{R} \times [-R, R]$:
\begin{align*}
\int_{-R}^{R} \int_{-\infty}^{\infty} |f(y) e^{ik(x-y)}| \, d\mathcal{L}^1(y) \, d\mathcal{L}^1(k) = \int_{-R}^{R} \|f\|_{L^1} \, d\mathcal{L}^1(k) = 2R \|f\|_{L^1} < \infty.
\end{align*}
Since $|f(y)e^{ik(x-y)}| = |f(y)|$ is independent of $k$, and $f \in L^1(\mathbb{R})$ by hypothesis, the product is integrable over $\mathbb{R} \times [-R, R]$. Fubini's theorem then permits interchanging the order of integration, bringing the $k$-integral inside. This converts the problem from Fourier coefficients to a convolution integral that we can analyse using the Dirichlet kernel.
[/guided]
[/step]
[step:Evaluate the inner integral to obtain the Dirichlet kernel]
For $y \neq x$, the inner integral evaluates to:
\begin{align*}
\int_{-R}^{R} e^{ik(x-y)}\,d\mathcal{L}^1(k) = \frac{e^{iR(x-y)} - e^{-iR(x-y)}}{i(x-y)} = \frac{2\sin(R(x-y))}{x - y}.
\end{align*}
For $y = x$, the integrand is $1$ and the integral equals $2R$. Define the Dirichlet kernel $D_R(u) = \frac{\sin(Ru)}{\pi u}$ for $u \neq 0$. Then:
\begin{align*}
S_R(x) = \int_{-\infty}^{\infty} f(y) \cdot \frac{\sin(R(x-y))}{\pi(x - y)}\,d\mathcal{L}^1(y) = \int_{-\infty}^{\infty} f(x - u)D_R(u)\,d\mathcal{L}^1(u),
\end{align*}
where we substituted $u = x - y$.
[/step]
[step:Split the integral and apply the Riemann-Lebesgue lemma]
Using the identity $\int_{-\infty}^{\infty} \frac{\sin(Ru)}{\pi u}\,d\mathcal{L}^1(u) = 1$ (which follows from the Dirichlet integral $\int_0^\infty \frac{\sin t}{t}\,d\mathcal{L}^1(t) = \frac{\pi}{2}$), write:
\begin{align*}
S_R(x) - \frac{1}{2}(f(x^+) + f(x^-)) &= \int_0^{\infty} \frac{f(x - u) - f(x^-)}{\pi u}\sin(Ru)\,d\mathcal{L}^1(u) \\
&\quad + \int_{-\infty}^{0} \frac{f(x - u) - f(x^+)}{\pi u}\sin(Ru)\,d\mathcal{L}^1(u),
\end{align*}
where we used $\frac{1}{2}f(x^-) = f(x^-)\int_0^\infty D_R(u)\,d\mathcal{L}^1(u)$ and similarly for $f(x^+)$.
[guided]
The idea is to subtract the limiting value from both sides and show the remainder vanishes. Since $\int_0^\infty D_R(u)\,d\mathcal{L}^1(u) = \frac{1}{2}$ and $\int_{-\infty}^0 D_R(u)\,d\mathcal{L}^1(u) = \frac{1}{2}$, we can write
\begin{align*}
S_R(x) - \frac{f(x^+)}{2} - \frac{f(x^-)}{2} = \int_0^\infty \frac{f(x-u) - f(x^-)}{\pi u}\sin(Ru)\,d\mathcal{L}^1(u) + \int_{-\infty}^0 \frac{f(x-u) - f(x^+)}{\pi u}\sin(Ru)\,d\mathcal{L}^1(u).
\end{align*}
The function $g_+(u) = \frac{f(x-u) - f(x^-)}{\pi u}$ for $u > 0$ has a well-defined limit as $u \to 0^+$ because $f$ is of bounded variation (the one-sided derivative exists). Similarly $g_-(u) = \frac{f(x-u) - f(x^+)}{\pi u}$ has a limit as $u \to 0^-$. Both $g_+$ and $g_-$ are absolutely integrable on their respective half-lines (near zero by the bounded variation hypothesis, and at infinity because $f \in L^1$). By the Riemann-Lebesgue lemma, $\int_0^\infty g_+(u)\sin(Ru)\,d\mathcal{L}^1(u) \to 0$ as $R \to \infty$, and similarly for the negative half-line integral.
[/guided]
[/step]
[step:Conclude that $S_R(x) \to \frac{1}{2}(f(x^+) + f(x^-))$ as $R \to \infty$]
Define $g_+: (0, \infty) \to \mathbb{C}$ by $g_+(u) = \frac{f(x - u) - f(x^-)}{\pi u}$ and $g_-: (-\infty, 0) \to \mathbb{C}$ by $g_-(u) = \frac{f(x - u) - f(x^+)}{\pi u}$.
Near $u = 0$: since $f$ is of bounded variation on every finite interval, the one-sided limits $f(x^{\pm})$ exist, and $|g_+(u)| \leq C$ in a neighbourhood of $0$ (by the bounded variation estimate on the difference quotient). At infinity: $|g_+(u)| \leq \frac{|f(x-u)| + |f(x^-)|}{|\pi u|}$, and $f \in L^1(\mathbb{R})$ ensures $g_+ \in L^1((0,\infty))$. The same argument applies to $g_-$.
By the Riemann-Lebesgue lemma applied to $g_+ \in L^1((0,\infty))$:
\begin{align*}
\int_0^\infty g_+(u)\sin(Ru)\,d\mathcal{L}^1(u) \to 0 \quad \text{as } R \to \infty.
\end{align*}
Similarly $\int_{-\infty}^0 g_-(u)\sin(Ru)\,d\mathcal{L}^1(u) \to 0$. Therefore:
\begin{align*}
\lim_{R \to \infty} S_R(x) = \frac{1}{2}\bigl(f(x^+) + f(x^-)\bigr).
\end{align*}
At points of continuity, $f(x^+) = f(x^-) = f(x)$, giving $f(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\tilde{f}(k)e^{ikx}\,d\mathcal{L}^1(k)$.
[/step]
