I recently learned of the result by Carleson and Hunt (1968) which states that if $f \in L^p$ for $p > 1$, then the Fourier series of $f$ converges to $f$ pointwise-a.e. Also, [Wikipedia](http://en.wikipedia.org/wiki/Convergence_of_fourier_series#Norm_convergence) informs me that if $f \in L^p$ for $1 < p < \infty$, then the Fourier series of $f$ converges to $f$ in $L^p$. Either of these results implies that if $f \in L^p$ for $1 < p < \infty$, then the Fourier series of $f$ converges to $f$ in measure.


My first question is about the $p = 1$ case. That is:



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> If $f \in L^1$, will the Fourier series of $f$ converge to $f$ in measure?
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I also recently learned that there exist functions $f \in L^1$ whose Fourier series diverge (pointwise) everywhere. Moreover, such a Fourier series may converge (Galstyan 1985) or diverge (Kolmogorov?) in the $L^1$ metric.


My second question is similar:



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> Do there exist functions $f \in L^1$ whose Fourier series converge pointwise a.e., yet diverge in the $L^1$ metric?
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> 
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(Notes: Here, I mean the Fourier series with respect to the standard trigonometric system. I am also referring only to the Lebesgue measure on [0,1]. Of course, if anyone knows any more general results, that would be great, too.)



Convergence of Fourier Series of $L^1$ Functions
The answer to your first question is no. There is an $L^1$ function with Fourier series not converging in measure.


In the Kolmogorov example of an $L^1$ function $f$ with a.e. divergent Fourier series, there is in fact a set of positive measure $E$ and a subsequence $n\_k$ such that for all $x$ in $E$, the absolute values of the partial sums $S\_{n\_k}$ of the Fourier series goes to infinity with $k$.


$$\forall x\in E,\ \ |S\_{n\_k}f(x)|\rightarrow \infty$$


This can be checked from the construction of $f$ in the original article of Kolmogorov, in its selected [works](http://books.google.fr/books?hl=fr&lr=&id=ikN59GkYJKIC&oi=fnd&pg=PR13&dq=diverge+partout+authornbsp%3Akolmogorov&ots=2OvYAnZNwa&sig=BXuHcDly4I-RQM1rRSuUCmWSZKY#v=onepage&q&f=false).


If $S\_nf$ converges in measure, then $S\_{n\_k}f$ must also converges in measure. This implies that there is a subsequence $n\_{k\_l}$ such that $S\_{n\_{k\_l}}f(x)$ converges a.e. $x$, a contradiction. 


