# An application of Fubini's theorem in almost everywhere convergence of the Marcinkiewicz criterion

The Marcinkiewicz criterion for convergence $a.e.$ states that for a function $f$ integrable on the unit circle ($f\in L^1(\mathbb{T})$) with $L^1$ modulus of continuity ($\omega_1(t)=\sup_{|h|\leq t}||\tau_hf-f||_{L^1(\mathbb{T})}$), then the partial sums of $f$ converge almost everywhere, $$s_jf\to f~a.e.~\text{on}~\mathbb{T}$$ as long as the average of the modulus of continuity is integrable near the origin, ie, $$\int^{\delta}_{0}\frac{\omega_1(t)}{t}dt<\infty .$$ The solution to this proceeds as follows:

• Set the Marcinkiewicz integral $$I(\beta)=\int^{\pi}_{-\pi}\left| \frac{f(\beta+t)-f(\beta)}{t} \right|dt.$$

• Take the average integral and apply Fubini's theorem $$\frac{1}{2\pi}\int^{\pi}_{-\pi}I(\beta)d\beta =\frac{1}{2\pi}\int^{\pi}_{-\pi}\int^{\pi}_{-\pi} \left| \frac{f(\beta+t)-f(\beta)}{t} \right|dtd\beta\leq 2\int^{\pi}_{0}\frac{\omega_1(t)}{t}dt<\infty,$$ ie, $I(\beta)<\infty~a.e.,~\beta\in\mathbb{T}.$ Apply the Dini criterion to get the result.

My questions are as follows:

$1.~$Based on the definition of the $L^1$ modulus of continuity, should there not be a supremum in the final integral $2\int^{\pi}_0\frac{\omega_1(t)}{t}dt?$

$2.~$Why can we apply Fubini's theorem? What conditions have presented themselves that allow us to manipulate the integrals and apply symmetry about the origin?

Ad 1) After applying Fubini, we get an integral $\int_{-\pi}^\pi \dots / |t|\, dt$, where the $\dots$ in the integrand are given by $$\int_{-\pi}^\pi |f(t +\beta) - f(\beta)|\,d\beta = \Vert \tau_{-t}f -f \Vert_1 \leq \omega_1 (|t|),$$ simply because of $|-t|\leq |t|$ and by definition of $\omega_1$.
Thus, we can estimate the integral by $$\int_{-\pi}^\pi \omega_1(|t|) /|t| \, dt = 2\int_0^\pi \omega_1 (t)/t\, dt.$$
Ad 2) You can always apply the Fubini-Tonelli theorem if the integrand is nonnegative and measurable (and the two measure spaces under consideration are $\sigma$ finite). It is easy to see that all this is satisfied here.