# Weak convergence in $L^p$ equivalent to pointwise almost everywhere convergence

Can weak convergence of a sequence $f_n\in L^p(\Omega, \mu)$ to some $f\in L^p(\Omega, \mu)$ be characterised as almost everywhere pointwise convergence?

Let us also assume the measure space is $\sigma$-finite.

I'm asking because for continuous functions on the unit interval, weak convergence is equivalent to pointwise convergence.

• Pointwise convergence alone does not imply weak convergence for continuous functions on the unit interval. You also need boundedness. – Daniel Fischer Feb 6 '16 at 15:59

The standard counterexample is to look at $L^2((0,1))$ with Lebesgue measure and take $f_n(x) = \sqrt{2} \sin(n \pi x)$. The functions $f_n$ are orthonormal in $L^2$, so by Bessel's inequality they converge weakly to 0. But pointwise, the sequence $\{f_n(x)\}$ diverges for every $x \in (0,1)$.