Vtyjkyuk

Let $H_n$ be the Fourier sequence of an $L^2$-function $f$, with $H_n$ being the $n$-th Fourier series of $f$.
If $f$ converges almost everywhere to a function $g$, then does it hold that $f = g$ almost everywhere?

It is true :

By the containment of $L^p$ spaces see David M. Bressoud's book (A Radical approach to Lebesgue's theory of Integration) on page 260 :

$ (int_a^b |H_n|^p dx)^frac 1p le (b-a)^frac (q-p)pq (int_a^b |H_n|^q dx)^frac 1q$
you can take p=1 and q=2

By Bessel's Inequality $int_a^b |H_n|^2 dx le int_a^b |f|^2 dx $

This implies $H_n $ and $f$ are in $L^1 bigcap L^2$ over a bounded interval [a b] and $H_n $ is uniformly integrable and since $H_n -f$ converges almost everywhere , one can apply Vitali Convergence theorem

It is also a known result in $L^2$ (Riesz-Fischer theorem) $limlimits_nmapsto infty int_a^b (H_n -f)^2 dx = 0$

and by the inequalities above :$limlimits_nmapsto infty int_a^b |(H_n -f)|dx = 0$

by Vitali convergence theorem $limlimits_nmapsto infty int_a^b |(H_n -f)| dx =int_a^b limlimits_nmapsto infty |(H_n -f)| dx= 0$

implying $limlimits_nmapsto infty (H_n -f) =0 $ almost everywhere

Therefore $ limlimits_nmapsto infty H_n = g = f$ almost everywhere