Sequence with bounded $L^p$ norm which converges in measure also converges weakly

Question

Let $\{f_n \}_{\mathbb{N}}$ be a bounded sequence in $L^p(X, \Sigma, \mu)$ (i.e, there exists $M > 0$ such that $\|f_n \|_{L^p} \leq M$ for all $n \in \mathbb{N}$), where $(X, \Sigma, \mu)$ is a measure space (if needed, one can add the hypothesis that $\mu$ is a finite measure, but I don't know if this is necessary) and $1 < p < +\infty$. Suppose $f_n$ converges in the measure $\mu$ to some $f \in L^p(X, \Sigma, \mu)$. I want to prove that $f_n$ also converges weakly to $f$.

I have only found proofs of this result assuming almost everywhere convergence in measure. I know convergence in measure implies convergence almost everywhere but only for a subsequence, which is not enough here. Riesz's representation gives us a characterization of weak convergence, but I haven't been able to use that successfully. I'd appreciate any help or any indication of references where this is proven.

Answer 1

Here is a short proof for the case where $\mu$ is finite, $f_n\xrightarrow{n\rightarrow\infty}f$ in measure, $(f_n,f)\subset L_p(\mu)$, $1<p<\infty$, $g\in L_{p'}$. Then for any $\varepsilon>0$, \begin{align} \int_X|f-f_n|gd\mu&=\int_{\{|f_n-f|\leq\varepsilon\}}|f-f_n|gd\mu+\int_{\{|f_n-f|>\varepsilon\}}|f-f_n|gd\mu\\ &\leq\|g\|_{p'}\|(f-f_n)\mathbb{1}_{\{|f_n-f|\leq\varepsilon\}}\|_p+\|f-f_n\|_p\|g\mathbb{1}_{\{|f_n-f|>\varepsilon\}}\|_{p'}\\ &\leq \|g\|_{p'}\varepsilon\mu(X)+2M\|g\mathbb{1}_{\{|f_n-f|>\varepsilon\}}\|_{p'} \end{align}

The last inequality follows by the fact that $f=\lim_kf_{n_k}$ pointwise $\mu$-a.s. along some subsequence, and then an application of Fatou's lemma.

There is $\delta>0$ such that $\int_A|g|^{p'}\,d\mu<\varepsilon^{p'}$ for all measurable set $A$ with $\mu(A)<\delta$. Now, the as $\mu(|f_n-f|>\varepsilon)\xrightarrow{n\rightarrow\infty}0$, there $N$ such that for $n\geq N$, $\mu(|f_n-f|>\varepsilon)<\delta$. Putting things together, we have that for $n\geq N$ $$\Big|\int_X(f-f_n)g\,d\mu\big|\leq\int_X|f_n-f||g|\,d\mu\leq \|g\|_{p'}\varepsilon\mu(M)+2M\varepsilon$$

It follows that $f_n\xrightarrow{n\rightarrow\infty} f$ in $\sigma(L_p,L_{p'})$-topology.

---

For $\mu(X)=\infty$, the statement of the OP follows from everywhere convergence case:

Theorem: In any measure space $(X,\mathcal{B},\mu)$, if $(f_n)\subset L_p$, $1<p<\infty$, is a bounded function that converges $\mu$-a.s. to some $f$, then $f_n$ converges to $f$ in $\sigma(L_p, L_{p'})$

To see this, suppose $f_n$ converges to $f$ in measure in the sense that $\mu(|f_n-f|>\varepsilon)\xrightarrow{n\rightarrow\infty}0$ for all $\varepsilon>0$. Any subsequence $f_{n'}$ has a subsequence $f_{n''}$ that converges to $f$ $\mu$-a.s. Hence $f_{n''}$ converges to $f$ in $\sigma(L_p,L_{p'})$. This would imply that the original sequence $f_n$ converges to $f$ in $\sigma(L_p,L_{p'})$. Indeed, if that were not the case, then there $g\in L_{p'}$ and $\varepsilon_0$, and a subsequence $f_{n'}$ such that
$$\big|\int(f_{n'}-f)g\,d\mu\big|\geq\varepsilon_0$$ But then $f_{n'}$ cannot have a subsequence $f_{n''}$ that convergence to $f$!