I'm working through the following problem in Royden (Chapter 5, exercise 7) and would like to understand the solution:

Let $E$ have finite measure, $\{f_n\}\to f$ in measure on $E$, and $g$ be a measurable function on $E$ that is finite $a.e.$ on $E$. Prove that $\{f_n\cdot g\}\to f\cdot g$ in measure.

$\newcommand{\N}{\mathbb{N}}$ Proof. Since $g$ is finite a.e. on $E$, $m(\bigcap_{n\in\N}\{x\in E\,\,|\,\,g(x)>n\})=0\implies \forall\epsilon>0\,\,\, \exists N_1\in\N$:$n\geq N_1 \implies m\{x\in E\,\,|\,\,|g(x)|>n\}<\epsilon$.

So, let $\epsilon>0$ be given. Then there exists $M\in\N$ so that $m\{x\in E\,\,|\,\,|g(x)|>M\}<\frac{\epsilon}{2}$.

Let $\eta>0$. Since $\{f_n\}\to f$ in measure on $E$, $\exists N\in\N :n\geq N\implies m\{x\in E\,\,|\,\,|f_n(x)-f(x)|>\frac{\eta}{2M}\}<\frac{\epsilon}{2}$. Letting $n\geq N$, notice that: \begin{align*} &\{x\in E\,\,|\,\,|(f_n\cdot g)(x)-(f\cdot g)(x)|>\frac{\eta}{2}\}\\ =&\{x\in E\,\,|\,\,|g(x)\cdot f_n(x)-g(x)\cdot f(x)|>\frac{\eta}{2}\}\\ =&\{x\in E\,\,|\,\,|g(x)|\cdot|f_n(x)-f(x)|>\frac{\eta}{2}\}\\ \subset &\{x\in E\,\,|\,\,|g(x)|>M_1\}\cup \{x\in E\,\,|\,\,|f_n(x)-f(x)|>\frac{\eta}{2M}\}.\\ \end{align*} We then infer that: \begin{align*} &m\{x\in E\,\,|\,\,|(f_n\cdot g)(x)-(f\cdot g)(x)|>\frac{\eta}{2}\}\\ \leq &m\{x\in E\,\,|\,\,|g(x)|>M_1\}+m\{x\in E\,\,|\,\,|f_n(x)-f(x)|>\frac{\eta}{2M}\}\\ < &\frac{\epsilon}{2}+\frac{\epsilon}{2}\\ =&\epsilon.\\ \end{align*} $\blacksquare$

Specifically, I'm having some trouble understanding why: $\{x\in E\,\,|\,\,|g(x)|\cdot|f_n(x)-f(x)|>\frac{\eta}{2}\} \subset \{x\in E\,\,|\,\,|g(x)|>M_1\}\cup \{x\in E\,\,|\,\,|f_n(x)-f(x)|>\frac{\eta}{2M}\}$.

Sorry if I'm being a bit handwavy. Any constructive feedback will be helpful!


1 Answer 1


For convenience we set $$F=\{x\in E:|g(x)||f_n(x)-f(x)|>\frac{\eta}{2}\}$$ $$F_1=\{x\in E:|g(x)||f_n(x)-f(x)|>\frac{\eta}{2}\text{ and }|g(x)|>M\}$$ $$F_2=\{x\in E:|g(x)||f_n(x)-f(x)|>\frac{\eta}{2}\text{ and }|g(x)|\leq M\}$$ $$F_3=\{x\in E:|g(x)|>M\}$$ $$F_4=\{x\in E:|f_n(x)-f(x)|>\frac{\eta}{2M}\}$$ It is trivial that $$F=F_1\cup F_2,$$ $$F_1\subseteq F_3.$$ On the other hand $|g(x)||f_n(x)-f(x)|>\frac{\eta}{2}$ together with $|g(x)|\leq M$ imply $|f_n(x)-f(x)|>\frac{\eta}{2M}$, therefore $$F_2\subseteq F_4.$$ Hence $$F\subseteq F_3\cup F_4.$$


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