Convergence of the product of an $L_1$-convergent function f and a point-wise a.e. convergent function g.

Question

Let $f_k,f$ be integrable functions with $f_k\to f$ in $L_1$, $g_k,g$ are measurable with $\sup_k\|g_k\|_{L_\infty}<\infty$ and $g_k\to g$ p.w. a.e.. Show that $f_kg_k\to fg$ in $L_1$.

Using Hölder's and Minkowsi's Inequalities I have been able to bound $\|f_kg_k-fg\|_{L_1}$ in the following way. $$ \begin{split} \|f_kg_k-fg\|_{L_1} & =\|f_kg_k-fg_k+fg_k-fg\|_{L_1}\\ \text{(Minkowski) }&\leq\|f_kg_k-fg_k\|_{L_1}+\|fg_k-fg\|_{L_1}\\ \text{(Hölder) }&\leq \|f_k-f\|_{L_1}\|g_k\|_{L^\infty}+ \|g_k-g\|_{L^\infty}\|f\|_{L^1}\\ &\leq \|g_k-g\|_{L^\infty}\|f\|_{L^1}. \end{split} $$ However, I am not sure about how to show that the the $\|g_k-g\|_{L^\infty}$ term goes to $0$. Did I go wrong somewhere and if not what approach can I use to show the convergence from here?

Answer 1

Instead of using Hölder in the second term, use the dominated convergence theorem, since you have pointwise convergence and

$ \left|f\left(g-g_{k}\right)\right| \leq \left(\lVert g\rVert_{L_{\infty}} + \sup_{k} \lVert g_{k} \rVert_{L_{\infty}}\right)\left|f\right| \in L_{1}. $

Notice that $\lVert g - g_{k}\rVert_{L_{\infty}}$ may not go to zero.

Answer 2

Here is a sketch of a proof based on a slight generalization of dominated convergence theorem:

Suppose $\phi_n\xrightarrow{n\rightarrow\infty}\phi$ and $\psi_n\xrightarrow{n\rightarrow\infty}\psi$ pointwise $\mu$-a.s., and that $|\phi_n|\leq \psi_n$ $\mu$-a.s. If $\lim_n\int \psi_n\,d\mu=\int \lim \psi_n\,d\mu$, then $\int|\phi-\phi_n|\,d\mu\xrightarrow{n\rightarrow\infty}0$

To apply this result to the probelm in the OP, notice that for any subsequence of $n'$ of $n\in\mathbb{N}$, there is a subsequence $n''$ such that $\lim_{n''\rightarrow\infty}g_{n''}f_{n''}=fg$ $\mu$ almost surely; by assumption, $$|f_ng_n|\leq \big(\sup_k\|g_k\|_\infty\big)|f_n|, \qquad n\in\mathbb{N}$$ and $$\lim_n\int|f_n|=\int\lim_n|f_n|=\int|f|$$ since $\big||f_n|-|f|\big|\leq |f_n-f|$. Therefore, $\lim_{n''}\int|g_{n''}f_{n''}-fg|\,d\mu=0$

A simple argument by contradiction shows that in fact $\lim_n\int|g_nf_n-fg|\,d\mu=0$. Indeed, suppose the contrary. Then there is $\varepsilon>0$ ans a subsequence $n'$ such that $$\inf_{n'}\int|g_{n'}f_{n'}-fg|\,d\mu\geq\varepsilon$$ The argument above shows that there is a subsequence $n''$ of $n'$ such that $\int|f_{n''}g_{n''}-fg|\,d\mu=0$, which is absurd!

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The general version of dominated convergence stated above has been discussed before in MSE. Here is a short proof of it:

With out loss of generality, we can assume that pointwise convergence and domination hold everywhere.

Clearly $|\phi|\leq \psi$ and so, $\phi\in L_1$. Since $\psi_n+\psi -|\phi_n-\phi|\geq 0$, from Fatou's lemma we obtain \begin{align} \int_\Omega 2\psi\,d\mu&\leq \liminf_n \int_\Omega(\psi_n+\psi-|\phi-\phi_n|)\,d\mu\\ &= 2\int_\Omega \psi\,d\mu +\liminf_n\left( -\int_\Omega|\phi-\phi_n|\,d\mu\right)\\ &= 2\int_\Omega \psi\,d\mu-\limsup_n\int_\Omega|\phi-\phi_n|\,d\mu. \end{align} Since $|\phi_n-f|\geq0$, $\limsup_n\int_\Omega|\phi-\phi_n|\,d\mu=0$. To conclude, notice that $\left|\int_\Omega(\phi_n -\phi)\,d\mu\right|\leq \int_\Omega|\phi_n -\phi|\,d\mu $.