A hint in Brezis' exercise 4.16.1

Question

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

Let $p \in [1, \infty)$ and $p'$ its Hölder conjugate. Let $f_n,f,g:\Omega \to \mathbb R$ be measurable functions such that $f_n,f \in L^p(\Omega)$ for all $n$. Assume $f_n \to f$ in the weak topology $\sigma(L^p, L^{p'})$ and $f_n \to g$ $\mu$-a.e. Then $f=g$ $\mu$-a.e.

1. Could you confirm if my attempt is fine?

2. Is there a way to bypass the use of exercise 3.4.1?

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Proof First, we need the following result, i.e.,

Brezis' exercise 3.4.1 Let $E$ be a Banach space and $(x_n) \subset E$ such that $x_n \to x$ in $\sigma(E, E^*)$. Then there exists a sequence $(y_n) \subset E$ such that $y_n \in \operatorname{conv} (\{x_n, x_{n+1}, \ldots\})$ for all $n$ and $y_n \rightarrow x$ in norm.

Let $(g_n)$ be a sequence given by above result for the pair $((f_n)_n, f)$. Then $g_n \in \operatorname{conv} (\{f_n, f_{n+1}, \ldots\})$ and $g_n \to f$ in $L^p$. There is a sub-sequence $\varphi$ of $\mathbb N$ such that $g_{\varphi (n)} \to f$ $\mu$-a.e.

Fix $\omega \in \Omega$ such that $f_n (\omega) \to g (\omega)$ and $g_{\varphi (n)} (\omega) \to f (\omega)$. Fix $\varepsilon >0$. There is $N$ such that for every $n >N$ $$ \begin{align} |f_n(\omega) - g (\omega)| &< \varepsilon. \end{align} $$

Fix $n>N$. There is $\psi (n) > \varphi (n)$ and a probability vector $(t_i)_{i=\varphi (n)}^{\psi (n)}$ such that $$ g_{\varphi (n)} = \sum_{i=\varphi (n)}^{\psi (n)} t_i f_i \quad \mu \text{-a.e.} $$

Then $$ |g_{\varphi (n)} (\omega) - g (\omega)| \le \sum_{i=\varphi (n)}^{\psi (n)} t_i |f_i (\omega) - g(\omega)| < \varepsilon. $$

Hence $g_{\varphi (n)} (\omega) \to g (\omega)$. It follows that $g(\omega) = f(\omega)$. This completes the proof.

Answer 1

@DavidMitra suggested an elegant proof in a comment. I reproduce it below.

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WLOG, we assume $f =0$ $\mu$-a.e. Assume the contrary that $\mu (\{g \neq 0\}) >0$. WLOG, we assume $\mu (\{g > 0\}) >0$. By continuity of measure from below, $$ \mu (\{g > 0\}) = \lim_n \mu (\{g > 1/n\}). $$

So there is $\varepsilon >0$ such that $\mu (B_\varepsilon)>0$ where $B_\varepsilon := \{g > \varepsilon\}$. Because $\mu$ is $\sigma$-finite, there is a subset $B$ of $B_\varepsilon$ such that $\delta := \mu(B)\in (0, \infty)$.

By Egorov theorem, there is $A \in \mathcal F$ such that $A \subset B$ and $\mu(A) < \delta$ and $f_n \to g$ uniformly on $C:=B \setminus A$. It follows that $\lim_n \int_C f_n = \int_C g \ge \varepsilon \mu(C)$. Notice that $\mu(C) = \delta - \mu(A) >0$. So $\lim_n \int_C f_n >0$.

We have $\mu(C) < \infty$, so $1_C \in L^{p'} (\Omega)$. By weak convergence of $(f_n)$, we have $\lim_n \int_C f_n = 0$, which is a contradiction. This completes the proof.

Answer 2

Here is another proof based on well known properties of $L_p$ spaces and the uniform boundedness principle. Throughout this posting we use $p'$ to denote the convex conjugate of $p$, i.e. $\frac1p+\frac1{p'}=1$.

Suppose $f_n\xrightarrow{n\rightarrow\infty}f$ in $\sigma(L_p,L_{p'})$ and that $f_n\xrightarrow{n\rightarrow\infty}g$ $\mu$.a.s. Since $f_n$ converges in $\sigma(L_p,L_{p'})$, an application of the uniform boundedness principle yields $M=\max(\|f\|_p,\sup_n\|f_n\|_p)<\infty$. Then, $\|g\|_p\leq M$ by Fatou's Lemma. For each $n\in\mathbb{N}$, define $g_n=f_n-g$.

Case $1<p<\infty$: Let $\phi\in L_{p'}$. Given $\varepsilon>0$ choose a simple function $\phi'$ such that $\|\phi-\phi'\|_{p'}<\frac{\varepsilon}{2M}$. By Hölder's inequality $$\Big|\int g_n\phi\,d\mu\Big|\leq\Big|\int g_n\phi'\,d\mu\Big|+\varepsilon$$ To prove the desired conclusion, it is enough to show that $\int g_n\phi'\,d\mu\xrightarrow{n\rightarrow\infty}0$ for any integrable simple function, and to prove this, it is enough to show that $\int g_n\mathbb{1}_A\,d\mu\xrightarrow{n\rightarrow\infty}0$ for all $A\in\mathcal{F}$ with $\mu(A)<\infty$. For such set $A$, by Egorov's theorem, there is $A'\subset A$ with $\mu(A\setminus A')<\big(\frac{\varepsilon}{2M}\big)^{p'}$ such that $g_n$ converges to $0$ uniformly on $A'$. Let $N\in\mathbb{N}$ be large enough so that $\sup_{x\in A'}\|g_n\|<\frac{\varepsilon}{2(\mu(A)+1)}$ for all $n\geq N$. It follows that \begin{align} \Big|\int_Ag_n\,d\mu\Big|&\leq\int_{A'}|g_n|\,d\mu+\int_{A\setminus A'}|g_n|\,d\mu\\ &\leq\sup_{x\in A'}\|g_n\|\mu(A)+M\mu^{1/{p'}}(A\setminus A')<\varepsilon \end{align} whenever $n\geq N$. Putting things together, we obtain that $f_n\xrightarrow{n\rightarrow\infty}g$ in $\sigma(L_p,L_{p'})$. It follows that $\int \phi f=\int \phi g$ for all $\phi\in L_{p'}$. By the uniqueness of the Riesz representation, it follows that $f=g$ $\mu$-a.s.

Case $p=1$: By assumption, for any $A\in \mathcal{F}$, $\lim_n\int_Af_n\,d\mu=\int_Af\,d\mu$. Since $(\Omega,\mathcal{F},\mu)$ is $\sigma$-finite, to show that $f=g$ $\mu$-a.s, it suffices to prove that $\int_Af=\int_Ag$ for all $A\in\mathscr{F}$ with $\mu(A)<\infty$. Given $\varepsilon>0$, there is $\delta>0$ such that for all $E\in\mathcal{F}$, $$\int_E|f-g|\,d\mu<\varepsilon/3 \qquad\text{if}\qquad \mu(E)<\delta$$ By Egorov's, there is $A'\subset A$ such that $\mu(A\setminus A')<\delta$ on where $g_n$ converges to $0$ uniformly. Choose $N\in\mathbb{N}$ such that $\sup_{x\in A'}|g_n(x)|<\frac{\varepsilon}{3(\mu(A)+1)}$. Then \begin{align} \Big|\int_Ag_n\,d\mu\Big|&\leq\int_{A'}|g_n|\,d\mu+\Big|\int_{A\setminus A'}f_n-g\,d\mu\Big|\\ &\leq \sup_{x\in A}|g_n(x)|\mu(A')+\int_{A\setminus A'}|f-g|\,f\mu+ \Big|\int_{A\setminus A'}f_n-f\,d\mu\Big|\\ &<\frac{\varepsilon}{3}+\Big|\int_{A\setminus A'}f_n-f\,d\mu\Big|\tag{1}\label{one} \end{align} The last term in \eqref{one} converges to $0$ as $n\rightarrow\infty$. The desired conclusion for $p=1$ follows.

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Comments:

1. $\sigma$-finitness does not play a role for $1<p<\infty$.
2. When $1<p<\infty$ we have in fact proven a stronger result: If $(f_n)\subset L_p$ is bounded (in $L_p$) and $f_n$ converges almost surely to some function $f$, then $f_n$ converges weakly to $f$.