pointwise convergence and boundedness in norm imply weak convergence

Question

I am contemplating over the following exercise (in which $E=[0,1]$):

Let $f_n$ be a sequence of functions in $L^p(E)$, $1<p<\infty$, which converge almost everywhere to a function $f$ in $L^p(E)$, and suppose that there is a constant $M$ such that $\|f_n\|_p\le M$ for all $n$. Then for each function $g$ in $L^q(E)$ (with $\frac1p+\frac1q=1$), we have $$\int_E fg=\lim_{n\to\infty}\int f_n g.$$

If measure of $E$ is finite, one can make use of Egoroff's theorem and find a set $A$ of small enough measure such that $f_n$ converges uniformly to $f$ on $E\setminus A$, and $\int_A|g|^q<\epsilon'^q$. Thus the difference $$ \left|\int_E fg - \int_E f_ng\right|\le\int_E |(f-f_n)g|=\int_A |(f-f_n)g|+\int_{E\setminus A}|(f-f_n)g| $$ $$ \le\text{[by Holder]}\le 2M\epsilon'+\|f-f_n\|_p\cdot\|g\|_q $$ can be made less than any desired $\epsilon$ for big enough $n$.

My question is: does this result also hold true if the measure of $E$ is infinite?

Answer 1

Just want to add another approach to this problem.

To show $\lim \int f_n g = \int fg$ for each $g\in L^q$, if $f_n$ are bounded, it suffices to show $\lim \int f_n \phi = \int f\phi$ for $\phi$ in a dense subset of $L^q$, say the space of simple functions in $L^q$ with compact support. Then your argument would apply perfectly.

Answer 2

Yes: consider for each integer $N$ the sets $S_N:=\{x,|g(x)|>N^{-1}\}$. Since $|g|^q$ is integrable, these sets have finite measure, and we are reduced to show the result when $E=\bigcup_NS_N$.

Then we use a $2\varepsilon$-argument: fix $\varepsilon>0$; we can find $N$ such that $\int_{E\setminus S_N}|g|^q<\varepsilon$. Then we do the same proof as the case $E$ of finite measure.

Answer 3

Yes. Here is a take on the matter https://www.researchgate.net/publication/241681011_Another_Proof_That_L_p_Bounded_Pointwise_Convergence_Implies_Weak_Convergence

For future reference, below a classical proof (see e.g. the excellent E. Hewitt and K. Stromberg, Real and Abstract Analysis, SpringerVerlag, 1975.) First we need the following Lemma: Suppose $(X,\mathcal{A},\mu)$ is a measure space. If $h \in L^1$ then $\forall \rho>0, \exists A \in \mathcal{A}: \mu(A)<\infty$ and $ \int_{A'}|h|d\mu < \rho$ where $A'$ is the complement of $A$ in $X$. Indeed, given $\rho>0$, there is a simple function $\sigma$ with $|\sigma|\leq |h|$ and $\parallel h-\sigma\parallel _1 < \rho$. Let $\alpha >0$ be the minimum of the non-zero absolute values taken on by $\sigma$, then we have $$\alpha\xi_A\leq |\sigma|\leq |h| $$ where $A=\left\{x \in X: |\sigma(x)|\neq 0 \right\}$ and $\xi_A$ is the indicator function of $A$ (notice that $A \in \mathcal{A}$). Therefore $$\alpha \mu(A)= \int_{A}\alpha\xi_A d\mu \leq \int_{A}|\sigma|d\mu \leq \int_{A}|h|d\mu < \infty$$ hence $\mu(A)<\infty$ and we have that $\int_{A'}|h|d\mu= \int_{A'}|h-\sigma + \sigma|d\mu\leq \int_{A'}|h-\sigma|d\mu + \int_{A'}|\sigma|d\mu\leq \parallel h-\sigma\parallel _1 <\rho $ since $|\sigma| = 0$ on $A'$.

We will also need Absolute Continuity of the Lebesgue Integral (this is a well known result): Suppose $(X,\mathcal{A},\mu)$ is a measure space and $f \in L^1(X,\mathcal{A},\mu)$. $\forall \epsilon>0,\exists \delta >0$ (only depending on $f$ and $\epsilon$): when $E\in \mathcal{A}$ and $\mu(E)<\delta$ then $$\int_{E}|f|d\mu< \epsilon$$

Finally we mention Egorov's Theorem: Let $(X,\mathcal{A},\mu)$ be a finite measure space (i.e. $\mu(X)<\infty)$ and $f_n$ and $f$ be measurable functions defined $a.e.$ with $f_n\to f \ a.e.$ Then $\forall \epsilon>0$ there exists $A \in \mathcal{A}$ such that $\mu(A')<\epsilon$ and $f_n \to f$ uniformly on $A$.

Now for the actual Theorem and proof. Suppose $1<p<\infty$ and $p'=\dfrac{p}{p-1}$. Let $f$ and $(f_n)_{n=1}^{\infty}$ be functions in $L^P(X,\mathcal{A},\mu)$ and suppose that $(\parallel f_n \parallel_p)_{n=1}^{\infty}$ is a bounded sequence of numbers. If $f_n \to f \ a.e.$ then $\int_{X}f_ngd\mu \to \int_{X}fgd\mu, \forall g \in L^{p'}(X,\mathcal{A},\mu)$.

Proof. Take $\alpha >0$ so that $\parallel f_n\parallel_p \leq \alpha, \forall n$. Fatou's lemma gives $$\parallel f \parallel_p^p =\int_{X}\lim_{n\to \infty}|f_n|^pd\mu=\int_{X} \varliminf_{n \to \infty} |f_n|^pd\mu \leq \varliminf_{n \to \infty} \int_{X} |f_n|^p d\mu=\varliminf_{n \to \infty}\parallel f_n \parallel_p^p \leq \alpha^p$$ so $$\parallel f \parallel_p \leq \alpha$$ In particular $\parallel f-f_n \parallel_p \leq \parallel f \parallel_p +\parallel f_n \parallel_p \leq 2\alpha $

Let $\epsilon > 0$ and $g \in L^{p'}(X,\mathcal{A},\mu).$

Absolute Continuity of the Lebesgue Integral on $|g|^{p'}$ guarantees a $\delta >0$ such that for all $E\in \mathcal{A}$ for which $\mu(E)<\delta$ $$2\alpha\left( \int_{E}|g|^{p'}d\mu\right)^{\dfrac{1}{p'}}< \dfrac{\epsilon}{3}$$ The Lemma on $|g|^{p'}$ gives us $A \in \mathcal{A}$ with $\mu(A)<\infty$ and $$2\alpha\left( \int_{A'}|g|^{p'}d\mu\right)^{\dfrac{1}{p'}}< \dfrac{\epsilon}{3}$$

We now apply Egorov's Theorem to the finite measure space $(A,\mathcal{A}_{|A},\mu_{|A})$ as subspace of $(X,\mathcal{A},\mu)$ with all the functions restricted on $A$. This gives us $B \in \mathcal{A}, B\subseteq A$ such that $\mu(A\cap B')< \delta$ and $f_n \to f$ uniformly on $B$. Hence there exists $n_0 \in \mathbb{N}$ such that $n \geq n_0$ implies $$|f(x)-f_n(x)|\left( \mu(B) \right)^{\dfrac{1}{p}}\parallel g \parallel_{p'}<\dfrac{\epsilon}{3}$$ for all $x \in B$.

We have $|\int_{X}fgd\mu-\int_{X}f_ngd\mu|=|\int_{X}(f-f_n)gd\mu|\leq \int_{X}|f-f_n||g|d\mu $ by Hölder's inequality.

Now $\int_{X}|f-f_n||g|d\mu = \int_{A\cap B'} |f-f_n||g|d\mu + \int_{A'}|f-f_n||g|d\mu+\int_{B}|f-f_n||g|d\mu \\ \leq \parallel f-f_n\parallel_p\left(\int_{A\cap B'}|g|^{p'}d\mu\right)^{\dfrac{1}{p'}}+ \parallel f-f_n\parallel_p\left(\int_{A'}|g|^{p'}d\mu\right)^{\dfrac{1}{p'}}+\left(\int_{B}|f-f_n|^pd\mu\right)^{\dfrac{1}{p}}\parallel g \parallel_{p'} $

by Minkowski's inequality. Putting everything together, if $n \geq n_0$, we have that $$|\int_{X}fgd\mu-\int_{X}f_ngd\mu|< \dfrac{\epsilon}{3} +\dfrac{\epsilon}{3}+\dfrac{\epsilon}{3}= \epsilon$$ Therefore, we have proved that $\int_{X}f_ngd\mu \to \int_{X}fgd\mu$