sequence in $L^1$ converging pointwise a.e., but not weakly.

Question

Find a $(X,\Sigma,\mu)$, a $\sigma$-finite measure space and a norm-bounded sequence $\{f_n\}$ in $L^1(X)$ that converges almost everywhere to $f$ but does not converge weakly to $f$.

Can you help me with this one?

Thanks.

Answer 1

This answers the original question, wich asked for an example of strong ($L^1$) convergence instead of pointwise a.e. convergence.

There is no such sequence because strong convergence implies weak convergence in normed spaces. For the other way around, look at $f_n(x) = \sin(nx) \to 0$ weakly, but not strongly in $L^2 [0,2\pi]$. $$\|f_n\|_{L^2(\mathbb T)}^2 = \int_0^{2\pi} \sin^2 nx \ \mathrm dx = \frac1n \int_0^{2\pi n} \sin^2 t \ \mathrm dt = \int_0^{2\pi} \sin^2 t \ \mathrm dt = \pi < \infty$$ Note that $\pi \not\to 0 = \|0\|_{L^2(\mathbb T)}$ is the reason why it doesn't converge weakly. Weak convergence here is a consequence of the Riemann-Lebesgue lemma (Fourier coefficients decay)