# pointwise convergence and boundedness in norm imply weak convergence

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?

• Are you sure that your proof when $E$ has finite measure works? What about the term $||f-f_{n}||_{p}.||g||_{q}$? How do you intend to remove that?
– Riju
Commented Jul 11, 2017 at 18:58
• @Riju $f_n$ converges uniformly to $f$ on a set of finite measure, hence it converges in $L_p$ norm as well. So the term $\|f-f_n\|_p$ can be made arbitrarily small, while $\|g\|_q$ is bounded. Commented Jul 11, 2017 at 23:03
• Yes. If you have the textbook Real Analysis, fourth edition by Royden and Fitzpatrick, see Proposition 9 of Section 8.2. Commented Apr 11, 2022 at 14:35
• It might be worth pointing out that the condition $f$ in $L^p(E)$ can be removed in the exercise. Commented Sep 30, 2022 at 20:13
• @user0 Theorem 12 actually. However their treatment is a complete shambles. Commented Nov 29, 2023 at 12:23

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.

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.

• Thanks! This is a very useful trick indeed! Commented Jul 13, 2013 at 11:04
• Why can you find such $N$? I've been trying to work this out but didn't come to any conclusion. Commented Jan 2, 2022 at 19:19

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 $$1 and $$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 $$\int_{X}f_ngd\mu \to \int_{X}fgd\mu$$