# Convergence in measure and boundness implies weak convergence in $L^p$

Let $$(f_n)_{n=1}^\infty$$ be a sequence in $$L^p ([0,1]), 1\leq p<\infty$$. Suppose that $$f_n\rightarrow f$$ in measure and that $$\sup\limits_{n\in\mathbb{N}} \|f_n\|<\infty$$. Show that $$f_n\rightarrow f$$ in the weak topology of $$L^p([0,1])$$.

As a reminder: a net $$\{g_\alpha\}_{\alpha\in I}$$ converges in the weak topology of $$X$$ if and only if $$\phi(g_\alpha)\rightarrow \phi(g)$$ for all $$\phi\in X^*$$.

We know convergence in measure and boundness in the $$L^p$$ norm gives us convergence in $$L^p$$. I think I could use the Dominated Convergence Theorem but I'm not sure how to proceed. I also found a proof of this Theorem but with the hypothesis $$f_n\rightarrow f$$ a.e., and they use Egorov's theorem, but I rather not use it.

I thank any suggestion you have.

1) For $$1< p< \infty$$. Let $$M = \sup \|f_n\|_{L^p}$$. Since $$f_n \to f$$ in measure, hence there is a subsequence $$f_{n_k}$$ which converges a.e. to $$f$$. By Fatou's lemma, we have $$f \in L^p$$ and $$\|f\|_{L^p} \leq M$$. We need to check $$\lim_{n\to \infty} \int_0^1 f_n g dx = \int_0^1 f g dx,\qquad\qquad (*)$$ for any $$g \in L^{p'}$$ with $$p' = p/(p-1)$$. Indeed, for any $$\epsilon >0$$, we have $$\lim_{n\to \infty} m(\{|f_n -f| > \epsilon\}) = 0$$ and $$|\int_0^1 (f_n -f)g dx| = |\int_{\{|f_n -f| >\epsilon\}} (f_n -f) g dx + \int_{\{|f_n -f| \leq \epsilon\}} (f_n -f) g dx,$$ and hence $$|\int_0^1 (f_n -f)g dx| \leq \int_{\{|f_n -f| >\epsilon\}} |f_n -f| |g| dx + \epsilon \int_{\{|f_n -f| \leq \epsilon\}} |g| dx.$$ By Holder inequality, we have $$\int_{\{|f_n -f| \leq \epsilon\}} |g| dx \leq \int_0^1 |g| dx \leq \|g\|_{L^{p'}},$$ and $$\int_{\{|f_n -f| >\epsilon\}} |f_n -f| |g| dx \leq (\int_{\{|f_n -f| >\epsilon\}} |f_n -f|^p dx)^{\frac1p}(\int_{\{|f_n -f| >\epsilon\}} |g|^{p'} dx)^{\frac1{p'}} \leq 2M (\int_{\{|f_n -f| >\epsilon\}} |g|^{p'} dx)^{\frac1{p'}}.$$ Therefore, it holds $$|\int_0^1 (f_n -f)g dx| \leq 2M (\int_{\{|f_n -f| >\epsilon\}} |g|^{p'} dx)^{\frac1{p'}} + \epsilon \|g\|_{L^{p'}}.\qquad\qquad (1)$$ We claim that $$\lim_{n\to \infty} \int_{\{|f_n -f| >\epsilon\}} |g|^{p'} dx = 0. \qquad\qquad (**)$$ Indeed, for any $$\delta >0$$, there exists $$A >0$$ such that $$\int_{\{|g| > A\}} |g|^{p'} dx < \delta$$. Hence $$\int_{\{|f_n -f| >\epsilon\}} |g|^{p'} dx \leq \int_{\{|f_n -f| >\epsilon\}\cap\{|g| > A\}} |g|^{p'} dx + \int_{\{|f_n -f| >\epsilon\}\cap\{|g| \leq A\}} |g|^{p'} dx \leq \delta + A^{p'} m(\{|f_n -f| > \epsilon\}).$$ Letting $$n \to \infty$$ and using $$m(\{|f_n -f| > \epsilon\}) \to 0$$, we get $$\limsup_{n\to \infty} \int_{\{|f_n -f| >\epsilon\}} |g|^{p'} dx \leq \delta,$$ for any $$\delta >0$$. Consequently, we get $$\limsup_{n\to \infty} \int_{\{|f_n -f| >\epsilon\}} |g|^{p'} dx \leq 0.$$ Evidently, we have $$\liminf_{n\to \infty} \int_{\{|f_n -f| >\epsilon\}} |g|^{p'} dx \geq 0$$. Hence, the claim $$(**)$$ holds. Letting $$n \to \infty$$ and using the claim $$(**)$$, we obtain $$\limsup_{n\to \infty} |\int_0^1 (f_n -f)g dx| \leq \epsilon \|g\|_{L^{p'}},$$ for any $$\epsilon >0$$. Therefore, it holds $$\limsup_{n\to \infty} |\int_0^1 (f_n -f)g dx| \leq 0.$$ Evidently, we have $$\liminf_{n\to \infty} |\int_0^1 (f_n -f)g dx| \geq 0$$. Hence, $$\lim_{n\to \infty} \int_0^1 (f_n -f) gdx =0,$$ which proves $$(*)$$.
2) For $$p =1$$, the conclusion does not holds. For example, consider $$f_n = n \chi_{(0,1/n)}$$, we have $$\|f_n\|_{L^1} =1$$ for any $$n$$ and $$f_n \to 0$$ in measure. However, we have $$1 \in L^\infty = (L^1)^*$$ and $$\int_0^1 f_n \, 1 dx = 1 \not\to 0.$$