Let $(f\_n)$ be a sequence bounded in $L^1 (a,b)$ such that there exists $f$ with $f\_n \to f$ a.e.


In which other senses is true that $f\_n \to f$? Is is true in $L^1(a,b)$? If there was weak convergence in $L^1$ then we would have strong convergence in $L^1$ (applying Césaro means).


Many thanks. Cheers


D



Convergence a.e and $L^1$ boundedness implies convergence in which sense?
There is convergence in some non-locally convex spaces, e.g. $L^p, 0 < p < 1$.


More generally, for any concave function $\Psi : \mathbb{R}\_+ \to \mathbb{R}\_+$, such that $\Psi(0) = 0$ and $\Psi(x) / x \to 0, x \to \infty$, we have $\intop \Psi(f(x) - f\_n(x)) dx \to 0$. Indeed, the boundedness of $f - f\_n$ in $L^1$ implies that $\Psi(f - f\_n)$ are [uniformly integrable](https://en.wikipedia.org/wiki/Uniform_integrability) (cf. Valee-Poussin's criterion), which is a sufficient condition for the convergence of integrals in Lebesgue's theorem.


Convergence in $L^1$, however, fails in general. The classical example is $f\_n = n \cdot \mathsf{1}[0, \frac{1}{n}]$, which converges to $0$ pointwise but not in $L^1$.


In fact, probably one of the most useful formulations of Lebesgue's theorem is that convergence in $L^1$ $\Leftrightarrow$ convergence in measure + uniform integrability.


