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

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.

Here is a short proof for the case where $$\mu$$ is finite, $$f_n\xrightarrow{n\rightarrow\infty}f$$ in measure, $$(f_n,f)\subset L_p(\mu)$$, $$1, $$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 $$(f_n)\subset L_p$$, $$1, is a bounded function that converges $$\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 $$f$$!

• Could you provide a reference to the claim "there is $\delta > 0$ such that ...?" Commented Dec 19, 2023 at 4:24
• That is a standard result names as absolute continuity of the integral. I am sure there are posting s in MSE, but I suggest you prove it yourself: suppose $f\in L_1(\mu)$ and define $f_M=-M\vee(f\wedge M)$. Then $|f_M|\leq M$ and $|f_M|\nearrow |f|$ ; hence $|f_M|$ converges to $|f|$ in $L_1$ by monotone convergence. Hence $\int_A|f|=\int_A(|f|-|f_M|)+\int_A|f_M|\leq \||f|-|f_M|\|_1+M\mu(A)$ ... Commented Dec 19, 2023 at 4:33
• I see. Thanks a lot! Commented Dec 19, 2023 at 4:37