$L^1$-convergence

Question

In my fourier analysis script we learnt that the fourier series of a function $f\in L^1[0,1]$ must not converge to $f$ in $L^1$, i.e. in general we have

$||S_n(f)-f||_{L^1[0,1]}\not\to 0$.

$S_n(f)$ are the symmetrical partial sums of the fourier series. Later, the author says that for all $f\in L^1[0,1]$ we have for all $x\in [0,1]$ that

$\int_{0}^xf(y)dy=\lim_{n\to\infty}\int_{0}^{x}S_n(f) dy$.

Isn't this for $x=1$ a contradiction?

Answer 1

No, $\left|\int (f_n-f) dx\right|\to 0$ (the convergence you have) does not imply $\int |f_n-f|dx\to 0$ ($L^1$ convergence).

Answer 2

No, it is not. Consider the following counterexample (not given by Fourier series, just an example in $L^1[0,1]$).

Let $f(x) = 0$, and $f_n(x) = 2^{n+1}1_{[2^{-n},2^{-(n+1)}]}(x)$. Then we have $f_n\to f$ pointwise, but $$\|f-f_n\|_{L^1} = 1$$ for all $n$.