= Solution
Define the one-dimensional <convolution kernel>
$$
k(t)=\begin{cases}\sin t/|t|^{3/2},&t\ne0,\\0,&t=0.\end{cases}
$$
It is a <Borel measurable function>, and $f(x,y)=k(x-y)g(y)$. The <function> $k(x-y)$ is <Borel measurable> by composition with the continuous difference map. Lifting the <Lebesgue measurable function> $g$ to the second coordinate preserves measurability for the <product sigma-algebra>, hence also for <Lebesgue measure> on $\mathbb R^2$. Products of <measurable functions> are measurable, so $f$ is <Lebesgue measurable>.
Near zero, $|\sin t|\leq|t|$; away from zero, $|\sin t|\leq1$. Consequently
$$
|k(t)|\leq\begin{cases}|t|^{-1/2},&0<|t|\leq1,\\|t|^{-3/2},&|t|>1.\end{cases}
$$
Both bounds are integrable on their respective regions by <power singularity integrability>. In particular,
$$
\boxed{\|k\|_1\leq2\int_0^1t^{-1/2}\,dt+2\int_1^\infty t^{-3/2}\,dt=4+4=8.}
$$
We use the following forms of the two product-integration theorems. On the product of <measure spaces> equipped with <sigma-finite measures>, <Tonelli's theorem> says that a nonnegative measurable <function> has equal product and iterated <integrals>, with infinity allowed. <Fubini's theorem> says that for an absolutely integrable measurable <function>, its sections are integrable for <almost every> fixed coordinate, and both orders of iterated <integration> equal the finite product <integral>. The <Lebesgue measure> on each factor is a <sigma-finite measure>, because the <real line> is a <countable union> of finite-length <intervals>.
Apply <Tonelli's theorem> first to $|f|$, before presuming finiteness. By <translation invariance of Lebesgue measure>,
$$
\begin{aligned}
\int_{\mathbb R^2}|f(x,y)|\,dx\,dy
&=\int_{\mathbb R}|g(y)|\left(\int_{\mathbb R}|k(x-y)|\,dx\right)dy\\
&=\|k\|_1\int_{\mathbb R}|g(y)|\,dy
\leq8\int_{\mathbb R}|g(y)|\,dy<\infty.
\end{aligned}
$$
Thus \b[$f$ is <Lebesgue integrable> on $\mathbb R^2$]. The <triangle inequality for an integral> now gives the requested estimate:
$$
\boxed{\left|\int_{\mathbb R^2}f(x,y)\,dx\,dy\right|
\leq\int_{\mathbb R^2}|f(x,y)|\,dx\,dy
\leq8\int_{\mathbb R}|g(y)|\,dy.}
$$
This is the <L1 norm identity for a translated kernel product> followed by a simple bound on the <convolution kernel>.
There is also a stronger signed conclusion. The integrable <convolution kernel> $k$ is an <odd function>, so $\int_{\mathbb R}k(t)\,dt=0$. Having established <absolute integrability>, we may apply <Fubini's theorem> to $f$ itself:
$$
\boxed{\int_{\mathbb R^2}f(x,y)\,dx\,dy
=\left(\int_{\mathbb R}k(t)\,dt\right)\left(\int_{\mathbb R}g(y)\,dy\right)=0.}
$$
Unlike part (a), the signed cancellation is justified by a finite <integral> of the <absolute value>. The constant eight is useful as an <L1 norm> bound even though this particular signed total vanishes.
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