Fourier transform inversion formula for $f\in L_1(\mathbb{R}^n)$ and Dini condition Let us define the Dini condition for a function $f\in L_1(-\infty,\infty)$, i.e. Lebesgue summable on $\mathbb{R}$, as

Given an $x\in\mathbb{R}$ there is a $\delta>0$ such that the Lebesgue integral $\int_{[-\delta,\delta]}|\frac{f(x+t)-f(x)}{t}|d\mu_t$ exists.

In all the post the integrals are to be intended as Lebesgue integrals. I know (p. 423, here, of Kolmogorov-Fomin's Элементы теории функций и функционального анализа) that, if $f\in L_1(-\infty,\infty)$ satisfies the Dini condition in $x\in\mathbb{R}$, then the inversion formula for the Fourier transform holds: $$f(x)=\frac{1}{2\pi}\lim_{N\to+\infty}\int_{[-N,N]}\Bigg(\int_{\mathbb{R}}f(t)e^{-i\lambda t)}d\mu_t\Bigg)e^{i\lambda x} d\mu_\lambda.$$
The same famous text proves (p. 437-438) a similar theorem for functions $f\in L_1(\mathbb{R}^n)$, under conditions that I do not fully understand, stated in the following way:

Let function $f(x_1,x_2,\ldots,x_n)$ ne integrable on the whole space $\mathbb{R}^n$ and let it satisfy the conditions:
  $$|f(x_1+t_1,x_2,\ldots,x_n)-f(x_1,x_2,\ldots,x_n)|\leq C|t_1|^a,$$ $$|f(x_1,x_2+t_2,\ldots,x_n)-f(x_1,x_2,\ldots,x_n)|\leq C(x_1)|t_2|^a,$$$$\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots$$$$|f(x_1,x_2,\ldots,x_n+t_n)-f(x_1,x_2,\ldots,x_n)|\leq C(x_1,x_2,\ldots,x_{n-
> 1})|t_n|^a,$$where $0\le a\le1,\quad\int_{\mathbb{R}}C(x_1)d\mu_{x_1}<\infty,\ldots,\quad\int_{\mathbb{R}}\ldots\int_{\mathbb{R}}C(x_1,\ldots,x_{n-1})d\mu_{x_1}\ldots d\mu_{x_{n-1}}<\infty.$
  Then $$(2\pi)^nf(x_1,x_2,\ldots x_n)=$$$$=\lim_{N_1\to\infty}\int_{-N_1}^{N_1}\Big(\ldots\lim_{N_{n-1}\to\infty}\int_{-N_{n-1}}^{N_{n-1}}\Big(\lim_{N_n\to\infty}\int_{-N_n}^{N_n}g(\lambda_1,\ldots,\lambda_n)e^{ix_n\lambda_n }d\mu_{\lambda_n}\Big)e^{ix_{n-1}\lambda_{n-1}}d\mu_{\lambda_{n-1}}...\ldots\Big)e^{ix_1\lambda_1}d\mu_{\lambda_1} $$

where the integrals are Lebesgue integrals and $g(\lambda_1,\ldots,\lambda_n):=\int_{\mathbb{R}}\ldots\int_{\mathbb{R}}f(x_1,\ldots,x_n)e^{-i\sum_{k=1}^nx_k\lambda_k}d\mu_{x_1}\ldots d\mu_{x_n}$.
The theorem is precisely stated in this way in the book (I've only changed some notation to clearly show the integrals are Lebesgue integrals), but I am not sure to understand what are $a$ and the $t_k$ in this wording. Does anybody passing by know the theorem and what the inequalities mean: must they hold for some $a\in[0,1]$ or for all $a\in[0,1]$? As to the $t_k$, $k=1,\ldots ,n$ must they satisfy the inequality for some $\delta>0$ such that $t_1,\ldots,t_n\le\delta$? 
The proof seems to use the fact that the function $f(-,x_2,\ldots ,x_n)$ as a function of $x_1$ satisfies the above defined Dini condition: does it? Why can we see that? In particular, I am not convinced that $a=0$ could guarantee the Dini condition ($\int_{0}^1 x^{-1}dx=+\infty$).
I think that my greatest problem is that I do not understand the conditions in the theorem's statement... Thank you so much for any explanation!
 A: The Riemann-Lebesgue lemma states that, for any $f \in L^{1}(\mathbb{R})$,
$$
                 \lim_{R\rightarrow\pm\infty}\int_{-\infty}^{\infty}f(t)e^{-iRt}\,dt = 0.
$$
This becomes the basis for a lot of basic pointwise convergence theorems. If $\mathcal{F}$ denotes the Fourier transform on $\mathbb{R}$, and $\mathcal{F}^{-1}$ its inverse transform, then
$$
             \mathcal{F}^{-1}(\chi_{[-R,R]}\mathcal{F}f)|_{x}-L=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{f(t)-L}{t-x}\sin(R(t-x))\,dt.
$$
The function $\frac{f(t)-L}{t-x}$ is absolutely integrable for $t \in (-\infty,x-\delta]\cup[x+\delta,\infty)$ whenever $\delta > 0$. If it is also integrable on $[x-\delta,x+\delta]$, then the Riemann-Lebesgue lemma gives
$$
           L = \lim_{R\rightarrow\infty}\mathcal{F}^{-1}(\chi_{[-R,R]}\mathcal{F}f)|_{x}.
$$
The Dini condition is nothing more than the additional assumption required to guarantee that $f \in L^{1}$ or $f\in L^{2}$ gives rise to a function $\frac{f(t)-L}{t-x}$ which is absolutely integrable in $t$ for a fixed $x$, where $L=f(x)$. One way to guarantee this condition is by assuming $f$ is $\alpha$-Holder continuous at $x$ with exponent $\alpha \in (0,1]$:
$$
                          |f(t)-f(x)| \le C|t-x|^{\alpha}.
$$
This condition is sufficient in order the Dini condition to hold, but not necessary, but Holder continuity is convenient, especially for Sobolev spaces.
The conditions stated in your theorem allow you to evaluate the full inverse Fourier transform on $\mathbb{R}^{n}$ by evaluating with iterated integrals. That's why the conditions stated are "staggered" in some sense. But they're set up so that the inversion integral can be evaluated using iterated integrals and limits. That is, if $B_{N}$ is the box in $\mathbb{R}^{n}$ described by $\{ x : |x_{k}| \le N_{k},\; 1 \le k \le n\}$, then
$$
        \mathcal{F}^{-1}(\chi_{B_{N}}(x)\mathcal{F}f)
           = \mathcal{F_{n}}^{-1}\chi_{[-N_n,N_n]}\mathcal{F_{n-1}}^{-1}\chi_{[-N_{n-1},N_{n-1}]}\cdots \mathcal{F_{1}}^{-1}\chi_{[-N_1,N_1]}\mathcal{F}f,
$$
where $\mathcal{F_{k}}^{-1}$ is the inverse Fourier transform in the $k$-th coordinate variable only.
Knowing that the inversion integral exists on the whole space allows you to evaluate by iterated limits, and the imposed conditions allow you to move the limits inside the transforms to the location where they can be evaluated using the Riemann-Lebesgue/Dini procedure described above for a single variable, one at a time.
