Convergence in measure and convergence of integral

Question

Let $g_n:\mathbb{R} \rightarrow \mathbb{R}$ are measurable, suppose $g_n \rightarrow 0$ in measure, and $\int g_n^2 dm\leq 1$ for all $n$. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a measurable function such that $\int f^2 dm< \infty$, prove that $\int fg_n dm \rightarrow 0$ as $n \rightarrow \infty$. Here $m$ stands for the Lebesgue measure.

My attempt is: since $g_n \rightarrow 0$ in measure, $\forall \epsilon>0, \exists N>0$ s.t. $m(\{|g_n|>\epsilon\})<\epsilon$. So by Cauchy -Schwarz inequality, $$\left|\int fg_n\right| \leq \int_{\{|g_n|>\epsilon\}}|fg_n|+\int_{\{|g_n|\leq\epsilon\}}|fg_n|\leq \left(\int_{\{|g_n|>\epsilon\}}f^2 \right)^{1/2}\left(\int_{\{|g_n|>\epsilon\}}g_n^2 \right)^{1/2} + \epsilon\int_{\{|g_n|\leq\epsilon\}}|f|$$

Here I got stuck, the $(\int_{\{|g_n|>\epsilon\}}g_n^2)^{1/2} \leq 1$ is obvious, but I don't know how to bound $(\int_{\{|g_n|>\epsilon\}}f^2 )^{1/2}$ and $\int_{\{|g_n|\leq\epsilon\}}|f|$.

Any hint or help would be greatly appreciated

Answer 1

I might add a further approach to the ones presented by @MW based on an application of uniform integrability and Vitali's convergence. In a $\sigma$-finite measure space $(X,\mathscr{A},\mu)$, one of the characterisations of uniform integrability for $L^1$ functions is the following: a subset of functions $\mathscr{F}\subseteq L^1$ is said to be uniformly integrable if we have $\sup_{v \in \mathscr{F}}\int |v|<\infty$ and for any decreasing sequence of measurable sets $A_k\downarrow \emptyset$ we have $\lim_k \sup_{v \in \mathscr{F}}\int_{A_k} |v|=0$. A reference for this is in Schilling's book. Now if we consider $(fg_n)_{n \in \mathbb{N}}$ and the fact that the Lebesgue measure is $\sigma$-finite we have that by CS inequality and the assumption $\|g_n\|_{L^2}\leq 1$ $$\sup_{n \in \mathbb{N}}\int |fg_n|\leq \sup_{n \in \mathbb{N}}\|f\|_{L^2}\|g_n\|_{L^2}\leq \|f\|_{L^2}<\infty$$ while we also have by dominated convergence (since $f \in L^2$): $$\lim_{k\to \infty}\sup_{n \in \mathbb{N}}\int_{A_k}|fg_n|\leq \lim_{k\to \infty}\|\mathbf{1}_{A_k}f\|_{L^2}=0$$ therefore our sequence is uniformly integrable. We also have that $fg_n\to 0$ in measure, this can be seen by $\mu(|fg_n|>\varepsilon)\leq \mu(|f|>k)+\mu(|g_n|>\varepsilon/k),\forall n,k \in \mathbb{N}$ then sending $n$ to infinity, and then $k$ to infinity. Therefore we can apply Vitali's convergence theorem and obtain $\int |fg_n|\to 0$.

Answer 2

Throughout, we assume with no loss of generality that $f$ and $g_n$ are nonnegative, since if we prove the theorem replacing $f$ and $g_n$ with $|f|$ and $|g_n|$ then the result follows.

[Update: the original proof used machinery that you might not have been exposed to and that on reflection is not really necessary, so I have included an alternative argument using only dominated convergence.]

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Proof via dominated convergence

Let $\delta>0$. Now, for sufficiently small $\epsilon>0$ and $\eta>0$ we have (by dominated convergence) $$\int_{\{f>\frac{1}{\epsilon}\}\cup \{f<\eta\}}f^2\leq \delta^2\text{.}$$ To see that this follows from dominated convergence, just observe that the sequence of functions $f^2\chi_{\{f>m\}\cup \{f<\frac{1}{m}\}}$ converges pointwise to $0$ and is dominated by $f^2$.

Therefore for all $g_n$ we have $$\int_{\{f>\frac{1}{\epsilon}\}\cup \{f<\eta\}}fg_n \leq \left(\int_{\{f>\frac{1}{\epsilon}\}\cup \{f<\eta\}}f^2\right)^{\frac{1}{2}}\cdot \left(\int_{\{f>\frac{1}{\epsilon}\}\cup \{f<\eta\}}g_n^2\right)^{\frac{1}{2}} \leq \delta\text{.}\tag{1}$$

Fixing any such $\eta$ and $\epsilon$, we also have your inequalities, but with the integrands restricted further to $\{\eta\leq f\leq\frac{1}{\epsilon}\}$, and also we will actually choose $N$ large enough so that for $n>N$ we have $m(\{|g_n|>\epsilon^3\})<\epsilon^3$, so that for large enough $n$ we get

\begin{align*} \int_{\{\eta\leq f\leq\frac{1}{\epsilon}\}} fg_n & \leq \left(\int_{\{g_n>\epsilon^3\}\cap \{\eta\leq f\leq\frac{1}{\epsilon}\}}f^2 \right)^{1/2}\left(\int_{\{g_n>\epsilon^3\}\cap\{\eta\leq f\leq\frac{1}{\epsilon}\}}g_n^2 \right)^{1/2} \\ &+ \epsilon^3\int_{\{g_n\leq\epsilon^3\}\cap\{\eta\leq f\leq\frac{1}{\epsilon}\}}f\tag{2}\\ & \leq \epsilon^{\frac{1}{2}}\cdot 1 + \frac{\epsilon^3}{\eta}\int_{\{g_n\leq\epsilon^3\}\cap\{\eta\leq f\leq\frac{1}{\epsilon}\}}f^2 \end{align*}

Then letting $\epsilon\to 0$ and combining (1) and (2) gives $$\overline{\lim_{n\to\infty}}\int fg_n \leq \overline{\lim_{n\to\infty}}\int_{\{f>\frac{1}{\epsilon}\}\cup \{f<\eta\}}fg_n + \overline{\lim_{n\to\infty}}\int_{\{\eta\leq f\leq\frac{1}{\epsilon}\}} fg_n\leq \delta\text{,} $$ and since $\delta$ was arbitrary this completes the proof.

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Proof via Banach–Alaoglu theorem

From the subsequence principle it is enough to show that every subsequence of $\int fg_n$ has a further subsequence which converges to $0$. But since every subsequence of $g_n$ satisfies the assumptions of the original sequence, this reduces our task to finding a single subsequence of the original sequence $\int fg_n$ that converges.

Now, by the weak-* compactness of the ball in $L^2(\mathbb R)$, there is a subsequence of $g_n$ (which we immediately rename as $g_n$ itself) that converges weak-* to some $g\in L^2(\mathbb R)$, i.e., $\int h g_n\to \int hg$ for each $h\in L^2(\mathbb R)$. (Here we are also using the characterization of the dual of $L^2$ via the Riesz representation theorem)

It therefore suffices to show that $g=0$ almost everywhere, as we may take $h=f$. Equivalently, we will show that $\int_A g=0$ for all $A$ with $m(A)<\infty$. To do this, we fix such $A$ and then simply repeat your calculation replacing $f$ with $\chi_A$:

\begin{align*} \int \chi_A g_n & \leq \int_{\{g_n>\epsilon\}}\chi_A g_n+\int_{\{g_n\leq\epsilon\}}\chi_A g_n\\ &\leq \left(\int_{\{g_n>\epsilon\}}\chi_A^2 \right)^{1/2}\left(\int_{\{g_n>\epsilon\}}g_n^2 \right)^{1/2} + \epsilon\int_{\{g_n\leq\epsilon\}}\chi_A\\ & \leq \epsilon^{1/2}\cdot 1 + \epsilon m(A) \end{align*} and so letting $\epsilon\to 0$ establishes that $\int g_n\chi_A\to 0$, whereby from weak convergence of $g_n$ we have $\int_A g = \int g\chi_A=0$, as desired.