= Solution
Use the <Fourier transform> convention $\widehat h(\xi)=\int h(x)e^{-2\pi ix\cdot\xi}\,dx$. The deleted-ball kernel need not be integrable at infinity, so first work with finite annuli. Write
$$
m_{\varepsilon,R}(\xi)=\int_{\varepsilon<|x|<R}K(x)e^{-2\pi ix\cdot\xi}\,dx.
$$
We will prove a uniform bound, convergence as $R\to\infty$, and then convergence as $\varepsilon\downarrow0$. This makes all Fourier-transform assertions precise for the <annularly cancelling singular convolution kernel>.
Fix $\xi\ne0$ and put $y=\xi/(2|\xi|^2)$, so $e^{-2\pi iy\cdot\xi}=-1$. For $R>r\geq4|y|$, let $\Omega=\{r<|x|<R\}$ and define
$$
I_{r,R}=\int_\Omega K(x)e^{-2\pi ix\cdot\xi}\,dx,\qquad J_{r,R}=\int_\Omega K(x-y)e^{-2\pi ix\cdot\xi}\,dx.
$$
The <Hörmander integral kernel condition> gives
$$
|I_{r,R}-J_{r,R}|\leq\int_{|x|>r}|K(x)-K(x-y)|\,dx.
$$
On changing variables $u=x-y$, the phase factor shows $J_{r,R}=-\int_{\Omega-y}K(u)e^{-2\pi iu\cdot\xi}\,du$. The symmetric difference between $\Omega$ and $\Omega-y$ lies in the two shells whose radii are within $|y|$ of $r$ and $R$. On each such shell, the kernel's size bound gives, by polar integration,
$$
\int_{s-|y|<|u|<s+|y|}|K(u)|\,du\leq A|\mathbb S^{d-1}|\log\frac{s+|y|}{s-|y|}\leq C_d A\frac{|y|}{s}\quad(s\geq4|y|).
$$
Therefore
$$
|I_{r,R}+J_{r,R}|\leq C_dA|y|(r^{-1}+R^{-1}),
$$
and adding the two comparisons proves the <oscillatory tail estimate for a singular kernel>:
$$
\boxed{|I_{r,R}|\leq\frac12\int_{|x|>r}|K(x)-K(x-y)|\,dx+C_dA\frac{|y|}{r}}.
$$
The integral on the right is at most $B$. Moreover, for fixed $\xi$, it tends to zero as $r\to\infty$ because the translated difference is absolutely integrable outside $2|y|$. The shell term also tends to zero. Thus the far Fourier integral is Cauchy and converges as its outer radius tends to infinity.
Set $r_0=4|y|=2/|\xi|$. If $\varepsilon<r_0$, annular cancellation allows subtraction of the constant phase on the near annulus:
$$
\begin{aligned}
\left|\int_{\varepsilon<|x|<r_0}K(x)e^{-2\pi ix\cdot\xi}\,dx\right|
&=\left|\int_{\varepsilon<|x|<r_0}K(x)(e^{-2\pi ix\cdot\xi}-1)\,dx\right|\\
&\leq2\pi A|\xi|\int_{|x|<r_0}|x|^{1-d}\,dx\leq C_d A.
\end{aligned}
$$
The same estimate applies to a shorter near annulus when $R<r_0$. The far annulus is bounded by $B/2+C_dA$, using the tail estimate; if $\varepsilon\geq r_0$, apply that estimate with $r=\varepsilon$. At $\xi=0$ the annular integral vanishes by cancellation. We have proved
$$
\sup_{\varepsilon,R,\xi}|m_{\varepsilon,R}(\xi)|\leq C_dA+B/2.
$$
For each fixed $\varepsilon>0$, the size bound makes $K_\varepsilon\in L^2$, and $K1_{\{\varepsilon<|x|<R\}}\to K_\varepsilon$ in $L^2$ as $R\to\infty$. By the <Plancherel theorem>, their Fourier transforms converge in $L^2$ to $\widehat K_\varepsilon$. The pointwise improper limits just established identify this transform, up to a null set, with
$$
m_\varepsilon(\xi)=\lim_{R\to\infty}m_{\varepsilon,R}(\xi).
$$
For example identification follows by selecting an almost-everywhere convergent subsequence from the $L^2$ convergence. Hence
$$
\boxed{\sup_{\varepsilon>0}\|\widehat K_\varepsilon\|_\infty\leq C_dA+B/2}.
$$
The near-annulus integral with the constant phase subtracted converges absolutely at zero, since its bound is proportional to $|x|^{1-d}$. The far integral has already converged. Therefore $m_\varepsilon(\xi)\to m(\xi)$ for every $\xi\ne0$, with the same uniform bound; the value at zero may be set arbitrarily.
For $f\in L^2$, the <convolution> $K_\varepsilon*f$ is absolutely convergent for each $x$ by <Cauchy-Schwarz inequality>. Its <Fourier transform> is $m_\varepsilon\widehat f$: first prove this for <Schwartz functions> and then approximate any $L^2$ input. The convolutions converge uniformly under this approximation by the same Cauchy-Schwarz bound, and the multiplier outputs converge in $L^2$ by their uniform operator bound, so these two definitions agree. Define the bounded <Fourier multiplier operator> $T$ by $\widehat{Tf}=m\widehat f$. <Dominated convergence theorem> and the <Plancherel theorem> now give
$$
\boxed{\|T_\varepsilon f-Tf\|_2^2=\int|m_\varepsilon-m|^2|\widehat f|^2\,d\xi\longrightarrow0}.
$$
For the final request, use a normalized <bump function>, $\int\phi=1$, so that its dilates form an <approximate identity>. Multiplying Fourier transforms gives
$$
\widehat{T(\phi_\varepsilon)*f}=m\widehat\phi(\varepsilon\xi)\widehat f=\widehat{\phi_\varepsilon*(Tf)}.
$$
These products are integrable by <Cauchy-Schwarz inequality>, since $T\phi_\varepsilon,f\in L^2$, and also define $L^2$ functions because $m$ and $\widehat\phi$ are bounded. Thus the <convolution> identity is justified even though $T\phi_\varepsilon$ need not be in $L^1$:
$$
T(\phi_\varepsilon)*f=\phi_\varepsilon*(Tf).
$$
The allowed standard <approximate identity> results, applied to $Tf\in L^2$, prove convergence to $Tf$ in norm and <almost everywhere>. This is the <mass normalization in mollified singular integrals>; it does not require proving almost-everywhere convergence of the sharp truncations $T_\varepsilon f$.
If “bump function” means merely smooth with compact support, the stated limit needs the unit-integral qualification. For arbitrary such $\phi$, the limit is $(\int\phi)Tf$ instead. This is also the almost-everywhere limit: at each <Lebesgue point> of $Tf$, subtract $(\int\phi)Tf(x)$ inside the convolution and bound the remaining local average by the boundedness and compact support of $\phi$. As a concrete counterexample without normalization, take $K(x)=1/(\pi x)$ in dimension one and $\phi=2\rho$ for a unit-integral <mollifier> $\rho$. This kernel has the required size, odd annular cancellation, and translation-difference integral $\log(3)/\pi$ on $\{|x|>2|y|\}$ for every $y\ne0$. Its limit operator is the nonzero <Hilbert transform>; by the <Hilbert-transform Fourier multiplier>, it is an $L^2$ isometry. For any nonzero $f\in L^2$ the mollified limit is $2Tf$, not $Tf$.
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