Brezis' exercise 4.22.4: convergence in weak topology $\sigma(L^1, L^\infty)$

Question

Let $(\Omega, \mathcal F, \mu)$ be a $\sigma$-finite measure space. I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,

Let $f, f_n \in L^1 (\Omega)$ be non-negative for all $n \in \mathbb N$. Assume that $\int_\Omega f_n \to \int_\Omega f$ and $\int_E f_n \to \int_E f$ for every $E \in \mathcal F$ with $\mu(E) < \infty$. Then $f_n \to f$ in the weak topology $\sigma(L^1, L^\infty)$.

Could you confirm if my attempt is fine?

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Proof Fix $B \in \mathcal F$ where $\mu(B)$ is not necessarily finite. It suffices to prove that $\int_B f_n \to \int_B f$. Fix $\varepsilon >0$.

Because $\mu$ is $\sigma$-finite, there is an increasing sequence $(\Omega_m) \subset \mathcal F$ such that $\Omega_m \uparrow \Omega$ and that $\mu(\Omega_m) < \infty$ for all $m$. Let $B_m := \Omega_m \cap B$. Let $C := \Omega \setminus B$ and $C_m := \Omega_m \cap C$. Then $B_m \uparrow B$ and $C_m \uparrow C$. By dominated convergence theorem, there is $M \in \mathbb N$ such that $\|1_{B_m} f - 1_B f\|_1 < \varepsilon$ and $\|1_{C_m} f - 1_C f\|_1 < \varepsilon$ for all $m \ge \overline m$. Let $B'_m := B \setminus B_m$ and $C'_m := C \setminus C_m$. Then $\|1_{B'_m} f\|_1 < \varepsilon$ and $\|1_{C'_m} f\|_1 < \varepsilon$ for all $m \ge \overline m$.

Clearly, $\mu(B_{\overline m})+\mu(C_{\overline m}) < \infty$, so $$ \lim_n \int_{B_{\overline m}} f_n = \int_{B_{\overline m}} f \quad \text{and} \quad \lim_n \int_{C_{\overline m}} f_n = \int_{C_{\overline m}} f. $$

Let $D_{\overline m} := B'_{\overline m} \cup C'_{\overline m}$. Notice that $D_{\overline m} = \Omega \setminus (B_{\overline m} \cup C_{\overline m})$, so $$ \lim_n \int_{D_{\overline m}} f_n = \int_{D_{\overline m}} f . $$

Because $f, f_n$ are non-negative, we get $\lim_n \|1_{D_{\overline m}} f_n \|_1= \|1_{D_{\overline m}} f \|_1$ and $$ \begin{align} \bigg | \int_{B'_{\overline m}} f_n - \int_{B'_{\overline m}} f \bigg | &\le \bigg | \int_{B'_{\overline m}} f_n \bigg | + \bigg |\int_{B'_{\overline m}} f \bigg | \\ &\le \bigg | \int_{D_{\overline m}} f_n \bigg | + \bigg |\int_{D_{\overline m}} f \bigg | \\ &= \|1_{D_{\overline m}} f_n \|_1 + \|1_{D_{\overline m}} f \|_1. \end{align} $$

Finally, $$ \begin{align} \lim_n \bigg | \int_B f_n - \int_B f \bigg | &\le \lim_n \bigg | \int_{B_{\overline m}} f_n - \int_{B_{\overline m}} f \bigg | + \lim_n \bigg | \int_{B'_{\overline m}} f_n - \int_{B'_{\overline m}} f \bigg | \\ &\le \lim_n \|1_{D_{\overline m}} f_n \|_1 + \|1_{D_{\overline m}} f \|_1 \\ &= 2 \|1_{D_{\overline m}} f \|_1 \\ &= 2 \|1_{B'_m} f\|_1 + 2 \|1_{C'_m} f\|_1 \\ &\le 4 \varepsilon. \end{align} $$

The claim then follows by taking the limit $\varepsilon \downarrow 0$.

Answer 1

Things can be stremlined:

Suppose $\Omega_n\nearrow\Omega$, $C=\Omega\setminus B$. Then $\Omega_n\cap B=B_n\nearrow B$ as $n\rightarrow\infty$. Notice that $B\setminus B_n=B\setminus \Omega_n=B\cap(\Omega\setminus\Omega_n)$

Given $\varepsilon>0$, choose $N$ such that $\int_{\Omega\setminus\Omega_N}|f|<\varepsilon$.

\begin{align} (f-f_n,\mathbb{1}_B)&=(f-f_n,\mathbb{1}_{B_N})+(f-f_n,\mathbb{1}_{B\setminus B_N})\\ &=(f-f_n,\mathbb{1}_{B_N}) + (f,\mathbb{1}_{B\setminus B_N})-(f_n\mathbb{1}_{B\setminus B_N}) \end{align}

Since $f,f_n$ are nonnegative

• $(f,\mathbb{1}_{B\setminus B_N})=\int_{B\setminus B_N}f\leq\int_{\Omega\setminus\Omega_N}f<\varepsilon$.
• Similarly, $(f_n,\mathbb{1}_{B\setminus B_N})\leq (f_n,\mathbb{1}_{\Omega\setminus \Omega_N})\xrightarrow{n\rightarrow\infty}(f,\mathbb{1}_{\Omega\setminus \Omega_N})<\varepsilon$ Putting things together $$\limsup_n|(f-f_n,\mathbb{1}_B)|\leq2\varepsilon $$