What difference do we have between almost everywhere convergence and convergence in $L^1$ norm?

Question

let $(X, \Sigma, \mu)$ be a measure space

$f_n \to f$ in $L^1$ norm means that : $f_n \in L^1$ for ever $n$ and $\int_X |f_n-f|d\mu \to 0$

almost everywhere convergence means that $f_n \to f$ pointwise except for a set of points $J$ which is negligible. meaning $\mu(J) = 0$

I'm still new to measure theory and I'd like to know what's in common or different beteween these two notions ? also if one implies the other or vice versa.

thanks !

Answer 1

You have already give the difference However there some couples of implication to each other

First if the convergence in norm holds true then the converseof the Lebesgue Convergence Dominated theorem says there is exist a subsequence $(f_{n_j})_j$ of $(f_n)_n$ that convergence almost everywhere to $f$

Second if the convergence almost everywhere holds true and there exists a function $g\in L^1$ such that $|f_n|\le g$ almost everywhere then The Lebesgue Convergence dominated theorem says the convergence in norm holds true too.

Answer 2

Let $f_n = 2^n \cdot 1_{\left(\frac{1}{2^n}, \frac{1}{2^{n-1}}\right]}$, $f_n$ pointwise converge to $f \equiv 0$. But $\int |f_n - f| d\mu = 1$, so no convergence in $L^1$.