= Solution
The defining integral is well-defined and bounded for every dimension: $|\mathcal Ff|\leq\|f\|_1$ and the multiplier is supported on a finite-<volume> ball. However \b[the asserted general-dimensional $L^1$ conclusion is false for the Euclidean radial cutoff printed in the original PDF]. The <nonintegrability of a radial triangular Fourier cutoff in three dimensions> is a source issue, not an omitted conjugate or OCR substitution.
For a direct counterexample in $n=3$, take the <Schwartz function> $f(x)=e^{-\pi|x|^2}$, whose <Fourier transform> is the same Gaussian. With $r=|x|$ and $a=2\pi r$, radial integration gives
$$
F_R(r)=\frac{4\pi}{a}\int_0^R \rho(1-\rho/R)e^{-\pi\rho^2}\sin(a\rho)\,d\rho.
$$
The amplitude $b(\rho)=\rho(1-\rho/R)e^{-\pi\rho^2}$ vanishes at both endpoints, and $b'(R)=-e^{-\pi R^2}$. Two integrations by parts, with a further one to bound the remainder, give
$$
F_R(r)=-\frac{4\pi e^{-\pi R^2}}{(2\pi r)^3}\sin(2\pi Rr)+O(r^{-4}).
$$
The integral $\int_1^\infty r^2|F_R(r)|\,dr$ diverges logarithmically for every fixed $R>0$. For example, restrict to the periodic subintervals on which $|\sin(2\pi Rr)|\geq1/2$; the error is smaller than half the leading term at sufficiently large $r$. Thus $F_R\notin L^1(\mathbb R^3)$, even for this smooth integrable $f$.
The familiar intended argument is valid in one dimension. There the inverse Fourier kernel is
$$
K_R(x)=R\left(\frac{\sin(\pi Rx)}{\pi Rx}\right)^2\geq0,\qquad\int K_R=1,
$$
and $F_R=K_R*f$ by <Fubini's theorem>. Its mass outside any fixed neighbourhood of zero tends to zero by scaling, so
$$
\|K_R*f-f\|_1\leq\int K_R(y)\|f(\cdot-y)-f\|_1\,dy\longrightarrow0
$$
by <continuity> of translations in $L^1$. For arbitrary dimension, replacing the radial cutoff by the product $\prod_{j=1}^n(1-|\xi_j|/R)_+$ gives a <tensor product> of these integrable kernels and the same <approximate identity> proof. These are explicit valid repairs, not claims that the PDF specified a product cutoff.
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