How do we reduce the inversion formula for the $n$-dimensional Fourier transform to the $1$-dimensional case using Fubini's theorem? Please consider the following theorem from Measures, Integrals and Martingales (2nd edition):


Question 1: The proof of the claim is straightforward, but I don't understand the remark in the "caution" paragraph. Isn't the right-hand side of (19.6) equal to the Bochner integral $\int\frac{e^{{\rm i}b\xi}-e^{{\rm i}a\xi}}{{\rm i}\xi}\hat\mu(\xi)\:\lambda({\rm d}\xi)$, where $\lambda$ denotes the Lebesgue measure on $\mathcal B(\mathbb R)$?
Moreover, I would like to understand how this identity can be generalized to the $n$-dimensional case. In the book, there is the following corollary (without proof), but I'm not sure how we obtain it:

Question 2: How exactly do we need to apply Fubini's theorem here in order to use the former result (19.6)? Do we need to apply (19.6) somehow to the finite measures $\mu_i$, where $$\mu_1(B):=\mu\left(B\times\times_{i=2}^n(a_i,b_i)\right)\;\;\;\text{for }B\in\mathcal B(\mathbb R)$$ and $\mu_2,\ldots,\mu_n$ are defined in the obvious analogous way?
 A: Lévy's theorem can be generalized to a multidimensional setting, however, the equivalent expressions in the left-hand-side of 19.1 are slightly more involved.  Here is a sketch of a proof:
If $[-\boldsymbol{T},\boldsymbol{T}]=\prod^n_{k=1}[-T_k,T_k]$ and
$[\boldsymbol{a},\boldsymbol{b}]=\prod^n_{k=1}[a_k,b_k]$ are the $n$--dimesional intervals, then
\begin{align}
\mu\big((\boldsymbol{a},\boldsymbol{b})\big)+L_\mu(\partial(\boldsymbol{a},\boldsymbol{b}))=
\frac{1}{(2\pi)^n}
\lim_{\boldsymbol{T}\rightarrow\infty}\int_{[-\boldsymbol{T},\boldsymbol{T}]}\int_{[\boldsymbol{a},\boldsymbol{b}]}
e^{-i\boldsymbol{y}\cdot\boldsymbol{t}}\widehat{\mu}(\boldsymbol{t})\,d\boldsymbol{y}\,d\boldsymbol{t}\tag{1}\label{inversion}
\end{align}
where $L_\mu$ is a weighted sum of the measure of the
$k$--dimensional hyperfaces $F_k$ ($k=0,\ldots,n-1$) of the $d$-dimensional cube $[\boldsymbol{a},\boldsymbol{b}]$
\begin{align}
L_\mu=\sum^{n-1}_{k=0}\frac{1}{2^{n-k}}\mu(F_k).
\end{align}
The derivation of~\eqref{inversion} is similar  as that of (19.1).
Define $\Psi_T(\theta)=\int^T_{-T}\frac{\sin \theta t}{t}\,dt$. It is easy to check that $\Psi_T(\theta)=2\operatorname{sign}(\theta)\,\int^{\theta T}_0\operatorname{sinc}(t)\,dt$, where $\operatorname{sinc}(t)=\frac{\sin t}{t}\mathbb{1}_{\mathbb{R}\setminus\{0\}}(t)+\mathbb{1}_{\{0\}}(t)$,  and
$\lim_{T\rightarrow\infty}\Psi_T(\theta)=\pi\operatorname{sign}(\theta)\pi$. Hence
$$\lim_{T\rightarrow\infty}\big(\Psi_T(\theta-a)-\Psi_T(\theta-b)\big)=2\pi\mathbb{1}_{(a, b)}(\theta) +\pi\mathbb{1}_{\{a,b\}}(\theta)$$
By Fubini's theorem
\begin{align}
\lim_{\boldsymbol{T}\rightarrow\infty}&\frac{1}{(2\pi)^n}
\int_{[-\boldsymbol{T},\boldsymbol{T}]}\int_{[\boldsymbol{a},\boldsymbol{b}]}
e^{-i\boldsymbol{y}\cdot\boldsymbol{t}}\widehat{\mu}(\boldsymbol{t})\,d\boldsymbol{y}\,d\boldsymbol{t}\\
&= \lim_{\boldsymbol{T}\rightarrow\infty}\frac{1}{(2\pi)^n}
\int_{[-\boldsymbol{T},\boldsymbol{T}]}\int_{[\boldsymbol{a},\boldsymbol{b}]}
e^{-i\boldsymbol{y}\cdot\boldsymbol{t}}\int_{\mathbb{R}^n}e^{i\boldsymbol{t}\cdot\boldsymbol{x}}\,\mu(d\boldsymbol{x})\,d\boldsymbol{y}\,d\boldsymbol{t}\tag{2}\label{two}\\
&=\lim_{\boldsymbol{T}\rightarrow\infty}\frac{1}{(2\pi)^n}
\int_{\mathbb{R}^n}\int_{[-\boldsymbol{T},\boldsymbol{T}]}\int_{[\boldsymbol{a},\boldsymbol{b}]}
e^{i(\boldsymbol{x}-\boldsymbol{y})\cdot\boldsymbol{t}} \,d\boldsymbol{y}\,d\boldsymbol{t}\,\mu(d\boldsymbol{x})\tag{3}\label{three}\\
&=\lim_{\boldsymbol{T}\rightarrow\infty}\frac{1}{(2\pi)^n}
\int_{\mathbb{R}^n}\int_{[-\boldsymbol{T},\boldsymbol{T}]}
\prod^n_{j=1}\frac{e^{i(x_j-a_j)t_j} - e^{i(x_j - b_j)t_j}}{it_j} \,d\boldsymbol{t}\,\mu(d\boldsymbol{x})\tag{4}\label{four}\\
&=\lim_{\boldsymbol{T}\rightarrow\infty}\frac{1}{(2\pi)^n}
\int_{\mathbb{R}^n}
\prod^n_{j=1}\int^{T_j}_{-T_j}\frac{e^{i(x_j-a_j)t_j} - e^{i(x_j - b_j)t_j}}{it_j} \,dt_j\,\mu(d\boldsymbol{x})\tag{5}\label{five}\\
&=\lim_{\boldsymbol{T}\rightarrow\infty}\frac{1}{(2\pi)^n}
\int_{\mathbb{R}^n}\Big(\prod^n_{k=1}\big(\Psi_{T_j}(x_j-a_j)-\Psi_{T_j}(x_j-b_j)\big)\Big)\,\mu(d\boldsymbol{x})\tag{6}\label{six}\\
&=\frac{1}{(2\pi)^n}
\int_{\mathbb{R}^n}\Big(\prod^n_{j=1}\big(2\pi\mathbb{1}_{(a_j,b_j)}+\pi\mathbb{1}_{\{a_j,b_j\}}\big)\Big)d\mu\tag{7}\label{seven}.
\end{align}
Here we have used Fubini-Tonelli's in passing from \eqref{two} to \eqref{three}, which works because the $n$-dimensional intervals involve are bounded (before passing to limit as $T_j\rightarrow\infty$ for all $1\leq j\leq n$) and $\mu$ is a finite measure. Fubini-Tonelli's used again to pass from \eqref{four} to \eqref{five}. To pass from \eqref{five} to \eqref{six} we use the fact that $\cos(x)=\cos(-x)$ and $\sin(-x)=-\sin(x)$. Finally, to pass from \eqref{six} to \eqref{seven} we use dominated convergence, which works because $\mu$ is a finite measure, and $\Psi_T$ is uniformly bounded in $T$, i.e., $\sup_{T>0}\|\Psi_T\|_\infty<\infty$.
The rest is a combinatoric argument. The measure of the lower dimensional faces can be obtained by similar methods.
For any  $J\subset\{1,\ldots,n\}$, let $\delta_{a_J}(d y_J)=\Pi_{j\in J}\delta_{a_j}$,
$\lambda_J(d y_J)=\Pi_{j\in J}dy_j$ and $[\boldsymbol{a}_J,\boldsymbol{b}_J]=\Pi_{j\in J}[a_j,b_j]$. If $k=\#J$ then $\mu(\{\boldsymbol{a}_J\}\times[a_{J^c},b_{J^c}])$ is given by
\begin{align*}
\lim_{\boldsymbol{T}\rightarrow\infty}&
\frac{1}{(2T)^k(2\pi)^{n-k}}
\int_{[-\boldsymbol{T},-\boldsymbol{T}]}\Big(\int_{\mathbb{R}^J\times
[\boldsymbol{a}_{J^c},\boldsymbol{b}_{J^c}]}
e^{-i\boldsymbol{y}\cdot\boldsymbol{t}}\widehat{\mu}(\boldsymbol{t})\,\delta_{\boldsymbol{a}_J}(d\boldsymbol{y}_J)\otimes
d\boldsymbol{y_{J^c}}\Big)\,d\boldsymbol{t}\\
&=\frac{1}{(2\pi)^{n-k}}\lim_{\boldsymbol{T}\rightarrow\infty}\int
\prod_J\frac{\sin T_j|x_j-a_j|}{T_j|x_j-a_j|}\prod_{J^c}\big(\Psi_{T_j}(x_j-a_j)
-\Psi_{T_j}(x_j-b_j)\big)\,d\boldsymbol{x}\\
&=\frac{1}{(2\pi)^{n-k}} \int\prod_J\mathbb{1}_{a_j}\prod_{J^c}\big(2\pi\mathbb{1}_{(a_j,b_j)}+\pi\mathbb{1}_{\{a_j,b_j\}}\big)\,d\mu.
\end{align*}
Remark 1: If $\mu\ll\lambda_n$, where $\lambda_n$ is Lebesgue's measure in $\mathbb{R}^n$, we have that $\mu(\partial [\boldsymbol{a},\boldsymbol{b}])=0$ since $\lambda_n(\partial [\boldsymbol{a},\boldsymbol{b}])=0$. Then \eqref{inversion} takes the form
$$
\int_{[\boldsymbol{a},\boldsymbol{b}]}f(\boldsymbol{x})\,d\boldsymbol{x} =\lim_{\boldsymbol{T}\rightarrow\boldsymbol{\infty}}\frac{1}{(2\pi)^n}\int_{[\boldsymbol{a},\boldsymbol{b}]}\int_{[-\boldsymbol{T},\boldsymbol{T}]}e^{-i\boldsymbol{y}\cdot\boldsymbol{t}}\widehat{\mu}(\boldsymbol{t})\,d\boldsymbol{t}\,d\boldsymbol{y}
$$
Remark 2: For any $\boldsymbol{a}\in\mathbb{R}^n$
$$\begin{align}
\frac{1}{\lambda_n([-\boldsymbol{T},\boldsymbol{T}])}\int_{[-\boldsymbol{T},\boldsymbol{T}]} e^{-i\boldsymbol{a}\cdot\boldsymbol{t}}\widehat{\mu}(\boldsymbol{t})\,d\boldsymbol{t}&=\int_{[-\boldsymbol{T},\boldsymbol{T}]}\int_{\mathbb{R}^n}e^{i(\boldsymbol{x}-\boldsymbol{a})\cdot\boldsymbol{t}}\mu(d\boldsymbol{x})\,d\boldsymbol{t}\\
&=\int_{\mathbb{R}^n}\prod^n_{j=1} \operatorname{sinc}(T_j|x_j-a_j|)\,\mu(d\boldsymbol{x})\\
&=\mu(\{a\})+\int_{\{a\}^c}\prod^n_{j=1} \operatorname{sinc}(T_j|x_j-a_j|)\,\mu(d\boldsymbol{x})
\end{align}
$$
Hence
$$\mu(\{\boldsymbol{a}\})=\lim_{\boldsymbol{T}\rightarrow\boldsymbol{\infty}}\frac{1}{\lambda_n([-\boldsymbol{T},\boldsymbol{T}])}\int_{[-\boldsymbol{T},\boldsymbol{T}]} e^{-i\boldsymbol{a}\cdot\boldsymbol{t}}\widehat{\mu}(\boldsymbol{t})\,d\boldsymbol{t}
$$
Remark 3: When $\widehat{\mu}\in L_1(\mathbb{R}^n)$, them $\mu\ll\lambda_n$ and $f:=\frac{d\mu}{d\lambda_n}$ is given by
$$f(\mathbf{y})=\frac{1}{(2\pi)^n}\int_{\mathbb{R}^n}e^{-i\boldsymbol{y}\cdot\boldsymbol{t}}\widehat{\mu}(\boldsymbol{t})\,d\boldsymbol{t}\qquad\lambda_n-a.s.$$
First notice that
$$\left|\frac{1}{\lambda_n([-\boldsymbol{T},\boldsymbol{T}])}\int_{[-\boldsymbol{T},\boldsymbol{T}]} e^{-i\boldsymbol{a}\cdot\boldsymbol{t}}\widehat{\mu}(\boldsymbol{t})\,d\boldsymbol{t}\right|\leq \frac{\|\widehat{\mu}\|_1}{\lambda_n([-\boldsymbol{T},\boldsymbol{T}])}\xrightarrow{\boldsymbol{T}\rightarrow\boldsymbol{\infty}}0
$$
This implies that $\mu$ is a continuous measure on $\mathbb{R}^n$. For any $n$-dimensional interval $[\boldsymbol{a},\boldsymbol{b}]$
$$\left|\int_{[\boldsymbol{a},\boldsymbol{b}]}e^{-i\boldsymbol{y}\cdot\boldsymbol{t}}\widehat{\mu}(\boldsymbol{t})\,d\boldsymbol{y}\right|\leq\lambda_n([\boldsymbol{a},\boldsymbol{b}])|\widehat{\mu}(\boldsymbol{t})|\in L_1(\mathbb{R}^n)$$
An application of Fubini's theorem and dominated convergence in \eqref{inversion} yields
$$\mu([\boldsymbol{a},\boldsymbol{b}])=\int_{[\boldsymbol{a},\boldsymbol{b}])}\frac{1}{(2\pi)^n}\int_{\mathbb{R}^n}e^{-i\boldsymbol{y}\cdot\boldsymbol{t}}\widehat{\mu}(\boldsymbol{t})\,d\boldsymbol{t}\,d\boldsymbol{y}
$$
The conclusion of Remark 3 follows.
